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**Stanislav Shkarin ^{*}**

Queens's University Belfast, Pure Mathematics Research Centre, University Road, Belfast, UK

- Corresponding Author:
- Stanislav Shkarin

Queens's University Belfast

Pure Mathematics Research Centre

University Road, Belfast, BT7 1NN, UK

**Tel:**+442890245133

**E-mail:**[email protected]

**Received date:** July 21, 2015; **Accepted date:** August 03, 2015; **Published date:** August 31, 2015

**Citation:** Shkarin S (2015) Existence Theorems in Linear Chaos. J Generalized Lie Theory Appl S1:009. doi:10.4172/generalized-theory-applications.S1-009

**Copyright:** © 2015 Shkarin S. This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.

**Visit for more related articles at** Journal of Generalized Lie Theory and Applications

Chaotic linear dynamics deals primarily with various topological ergodic properties of semigroups of continuous linear operators acting on a topological vector space. In this survey paper, we treat questions of characterizing which of the spaces from a given class support a semigroup of prescribed shape satisfying a given topological ergodic property.

In particular, we characterize countable inductive limits of separable Banach spaces that admit a hypercyclic operator, show that there is a non-mixing hypercyclic operator on a separable infinite dimensional complex Fréchet space X if and only if X is non-isomorphic to the space ω of all sequences with coordinatewise convergence topology. It is also shown for any k ∈ N, any separable infinite dimensional Fréchet space X non-isomorphic to ω admits a mixing uniformly continuous group {Tt}t∈Cn T of continuous linear operators and that there is no supercyclic strongly continuous operator semigroup {Tt}t≥0 on ω. We specify a wide class of Fréchet spaces X, including all infinite dimensional Banach spaces with separable dual, such that there is a hypercyclic operator T on X for which the dual operator T′ is also hypercyclic. An extension of the Salas theorem on hypercyclicity of a perturbation of the identity by adding a backward weighted shift is presented and its various applications are outlined.

Hypercyclic **operators**; Mixing semigroups; Backward weighted shifts; Bilateral weighted shifts

Unless stated otherwise, all vector spaces in this article are over the field , being either the field of complex numbers or the field of real numbers, all topological spaces *are assumed* to be Hausdorff and all vector spaces *are assumed* to be *non-trivial*. As usual, is the set of integers, is the set of non-negative integers, is the set of positive integers and is the set of non-negative real numbers. Symbol L(X, Y) stands for the space of continuous linear operators from a topological vector space X to a topological vector space Y. We write L(X) instead of L(X, X) and X′ instead of L(X, ). For each T ∈ L(X), the dual operator T′: X′ → X′ is defined as usual: *(T′ f)(x)=f (Tx)* for f ∈ X′ and x ∈ X. By a *quotient* of a topological vector space X we mean the space X/Y, where Y is a closed linear subspace of X. We start by recalling some definitions and facts.

**Notation and definitions**

A topological vector space is called *locally convex* if it has a base of neighborhoods of zero consisting of convex sets. Equivalently, a topological vector space is locally convex if its topology can be defined by a family of seminorms. For brevity, we say *locally convex space* for a locally convex topological vector space. A subset B of a topological vector space X is called *bounded* if for any neighborhood U of zero in X, a scalar multiple of U contains B. We say that τ is a *locally convex topology* on a vector space X if (X, τ) is a locally convex space. If X is a vector space and Y is a linear space of linear functionals on X separating points of X, then the weakest topology on X, with respect to which all functionals from Y are continuous, is denoted σ(X, Y). The elements of X can be naturally interpreted as linear functionals on Y, which allows one to consider the topology σ(Y, X) as well. If is a family of bounded subsets of (Y, σ(Y, X)), whose union is Y, then the seminorms

define the *topology on X of uniform convergence on sets of the family* . The topology on X of uniform convergence on all bounded subsets of (Y, σ(Y, X)) is called the *strong topology* and denoted β(X, Y). The topology of uniform convergence on all compact convex subsets of (Y, σ(Y, X)) is called the *Mackey topology* and is denoted τ (X, Y). According to the Mackey-Arens theorem, for a locally convex space (X, τ) and a space Y of linear functionals on X, the equality Y=X′ holds if and only if σ(X, Y) ⊆ τ ⊆ τ (X, Y). We say that a locally convex space (X, τ) *carries a weak topology* if τ coincides with σ(X, Y) for some space Y of linear functionals on X, separating points of X. If X is a locally convex space, we write X_{β} for (X′, β(X, X′)), X_{τ} for (X, τ (X, X′)) and X_{σ} for (X, σ(X, X′)). Similarly we denote (X, β(X′, X)) by X_{β}′, (X′, τ (X′, X)) by X_{τ}′ and (X′, σ(X′, X)) by X_{σ}′. An -space is a complete metrizable topological vector space. A locally convex -space is called a Fréchet space. If {X_{α}: α ∈ A} is a family of locally convex spaces, then their (locally convex) *direct sum* is the algebraic direct sum of the vector spaces X_{α} endowed with the strongest locally convex topology, which induces the original topology on each X_{α}. Let be a sequence of vector spaces such that X_{n} is a subspace of X_{n+1} for each n ∈ Z_{+} and each X_{n} carries its own locally convex topology τ_{n} such that τ_{n} is (maybe nonstrictly) stronger than the topology τ_{n+1}|_{Xn}. Then the *inductive limit* of the sequence {X_{n}} is the space endowed with the strongest locally convex topology τ such that τ|_{Xn} ⊆ τ_{n} for each n ∈. In other words, a convex set U is a neighborhood of zero in X if and only if U ∩ X_{n} is a neighborhood of zero in X_{n} for each n ∈. An LB-*space* is an inductive limit of a sequence of Banach spaces. An LB_{s}-*space* is an inductive limit of a sequence of separable Banach spaces. We use symbol φφ to denote the locally convex direct sum of countably many copies of the one- dimensional space and the symbol ω to denote the product of countably many copies of . Note that φ is a space of countable algebraic dimension and carries the strongest locally convex topology (=any seminorm on φ is continuous). We can naturally interpret ω as the space of all sequences with coordinatewise convergence topology. Clearly ω is a separable *Fréchet* space. Recall also that if X is a locally convex space and A ⊂ X′, then A is called *uniformly equicontinuous* if there exists a neighborhood U of zero in X such that |f (x)| ≤ 1 for any x ∈ U and f ∈ A.

Let T be a continuous linear operator on a topological vector space X. A vector x ∈ X is called a *cyclic vector* for T if the linear span of the orbit O(T, x)={T^{n}: n ∈ } of x is dense in X. The operator T is called cyclic if T has a cyclic vector. Recall also that for n ∈ , T is called n-cyclic if there are vectors x_{1},...,x_{n} ∈ X such that the linear span of the set {T^{n}x_{j} : n ∈ , 1 ≤ j ≤ n} is dense in X. Obviously, 1-cyclicity coincides with cyclicity. We say that T is *multicyclic* if it is n-cyclic for some n ∈ .

Let X and Y be topological spaces and {T_{a}: a ∈ A} be a family of continuous maps from X to Y. An element x ∈ X is called *universal* for this family if the orbit {T_{a}x: a ∈ A} is dense in Y and {T_{a}: a ∈ A} is said to be universal if it has a universal element. We say that a family {T_{n} : n ∈ } is *hereditarily universal* if any its infinite subfamily is universal. An *operator semigroup* on a topological vector space X is a family {T_{t}}_{t}∈A of elements of L(X) labeled by elements of an abelian monoid A (monoid is a semigroup with identity) and satisfying T_{0}=I, T_{s+t}=T_{t}T_{s} for any t, s ∈ A (unless stated otherwise, we use additive notation for the operation on A). A *norm* on A is a function | • | : A → [0, ∞) satisfying |na|=n|a| and |a + b| ≤ |a| + |b| for any for any n ∈ and a, b ∈ A. An abelian monoid equipped with a norm will be called a normed semigroup. We will be mainly concerned with the case when A is a closed (additive) subsemigroup of containing 0 with the norm |a| being the Euclidean distance from a to 0. In the latter case we consider A to be equipped with topology inherited from and we say that an operator semigroup {T_{t}}_{t∈A} is *strongly continuous* if the map from A to X is continuous for any x ∈ X. We say that an operator semigroup {T_{t}}_{t∈A} is *uniformly continuous* if there exists a neighborhood U of zero in X such that for any sequence of elements of A converging to t∈A, T_{tn}x converges to T_{t}x uniformly on U. Clearly, uniform continuity is strictly stronger than strong continuity. It is also worth mentioning that many authors use the term ’uniformly continuous semigroup’ for semigroups satisfying the weaker of uniform convergence of T_{tn}x to T_{t}x on any bounded subset of X.

If A is a normed semigroup and {T_{t}}_{t∈A} is an operator semigroup on a topological vector space X, then we say that {T_{t}}_{t∈A} is (topologically) *transitive* if for any non-empty open subsets U, V of X, the set {|t| : t ∈ A, Tt(U)∩V ≠∅} is unbounded. We say that {T_{t}}_{t∈A} is (topologically) *mixing* if for any non-empty open subsets U, V of X, there is r=r(U,V) > 0 such that T_{t}(U)∩V≠∅ provided |t| > r. We also say that {T_{t}}_{t∈A} is hypercyclic (respectively, supercyclic) if the family {T_{t}:t ∈ A} (respectively, {zT_{t}: z ∈ , t ∈ A}) is universal. {T_{t}}_{t∈A} is said to be hereditarily hypercyclic (respectively, *hereditarily supercyclic*) if for any sequence of elements of A such that |t_{n}| → ∞, the family {T_{tn}:n ∈ } (respectively, {zT_{tn}:z ∈ , n ∈ }) is universal. A continuous linear operator T acting on a topological vector space X is called *hypercyclic, supercyclic, hereditarily hypercyclic, hereditarily supercyclic, mixing or transitive* if the semigroup has the same property. It is worth noting that our definition of a hereditarily hypercyclic operator follows Ansari [1], while in the terminology of references [2,3], the same property is called ’hereditarily hypercyclic with respect to the sequence n_{k}=k of all nonnegative integers’. Hyper- cyclic and supercyclic operators have been intensely studied during last few decades, [4-6] and references therein. Clearly mixing implies transitivity and hereditary hypercyclicity (respectively, hereditary supercyclicity) implies hypercyclicity (respectively, supercyclicity). Recall that a topological space X is called a *Baire space* if the intersection of countably many dense open subsets of X is dense in X. According to the classical Baire theorem, complete metric spaces are Baire.

**Proposition 1.1.** *Let X be a topological vector space, A be a normed semigroup and S={T _{a}}_{a∈A} be an operator semigroup on X. Then*

*(1.1.1) if is hereditarily hypercyclic, then is mixing.*

*If additionally X is Baire separable and metrizable, the converse implication holds:*

(1.1.2) if is mixing, then is hereditarily hypercyclic.

The above proposition is a combination of well-known facts, appearing in the literature in various modifications. It is worth noting that a similar statement holds for hypercyclicity and transitivity under certain natural additional assumptions. One can also write down and prove a supercyclicity analogue of the above proposition. In the next section we shall prove Proposition 1.1 for sake of completeness. It is worth noting that for any subsemigroup A0 of A, not lying in the kernel of the norm, {T_{t}}_{t∈A0} is mixing if {T_{t}}_{t∈A} is mixing. In particular, if {T_{t}}_{t∈A} is mixing, then T_{t} is mixing whenever |t| > 0.

The question of existence of supercyclic or hypercyclic operators or semigroups on various types of topological vector spaces was intensely studied. There are no hypercyclic operators on any finite dimensional topological vector space and there are no supercyclic operators on a finite dimensional topological vector space of real dimension > 2. These facts follow, for instance from the main result of reference [7]. Herzog [8] demonstrated that there is a supercyclic operator on any separable infinite dimensional Banach space. Later Ansari [9] and Bernal-Gonzáles [10], answering a question raised by Herrero, showed independently that for any separable infinite dimensional Banach space X there is a hypercyclic operator T ∈ L(X). Using the same idea as in reference [9], Bonet and Peris [11] proved that there is a hypercyclic operator on any separable infinite dimensional *Fréchet* space and demonstrated that there is a hypercyclic operator on an inductive limit X of a sequence X_{n} for n ∈ of separable Banach spaces provided there is n ∈ for which X_{n} is dense in X. Grivaux [3] observed that hypercyclic operators T constructed in references [9-11] are in fact mixing and therefore hereditarily hypercyclic. They actually come from the same source. Namely, according to Salas [12] an operator of the shape I + T, where T is a backward weighted shift on , is hypercyclic. Virtually the same proof as in reference [12] demonstrates that these operators are in fact mixing. Moreover, all operators constructed in the above papers, except for the ones acting on ω, are hypercyclic because of a quasisimilarity with one of the operators of the shape identity plus a backward weighted shift. The same quasisimilarity transfers the mixing property as effectively as it transfers hypercyclicity. A similar idea was used by Bermúdez, Bonilla and Martinón [13] and Bernal- González and Grosse-Erdmann [14], who have demonstrated that any separable infinite dimensional Banach space supports a hypercyclic strongly continuous semigroup {T_{t}}_{t∈R+}. Bermúdez, Bonilla, Conejero and Peris [15] have shown for any separable infinite dimensional complex Banach space X, there exists a mixing strongly continuous semigroup {T_{t}}_{t∈Π} with Π={z ∈ : Re z ≥ 0} such that the map is holomorphic on the interior of Π. As a matter of fact, one can easily see that the semigroup constructed in reference [15] extends to a holomorphic mixing group . Finally, Conejero [16] proved that any separable infinite dimensional complex *Fréchet* space nonisomorphic to ω supports a uniformly continuous mixing operator semigroup {T_{t}}_{t∈R+}.

The following theorem extracts the maximum of the method both in terms of the class of spaces and semigroups. Although the general idea remains the same, the proof requires dealing with a number of technical details of various nature. In particular, we will prove and apply a multi- operator version of the Salas theorem. For brevity we shall introduce the following class of locally convex spaces.

**Definition 1.2.** We say that a sequence of elements of a topological vector space X is an -sequence if x_{n} → 0 in X and the series converges in X for each a ∈ .

We say that a locally convex space X belongs to the class M if its topology is not weak and there exists an -sequence in X with dense span.

**Theorem 1.3.** *Let X ∈ M. Then for any k ∈ , there exists a hereditarily hypercyclic (and therefore mixing) uniformly continuous operator group . Moreover, if = the map from ^{k} to X is holomorphic for each x ∈ X.*

Since for any hereditarily hypercyclic semigroup and any non-zero t ∈^{k}, the operator Tt is hereditarily hypercyclic, we have the following corollary.

**Corollary 1.4.** Let X ∈ M. Then there is a hereditarily hypercyclic (and therefore mixing) operator T ∈ L(X).

**Remark 1.5.** It is easy to see that if X ∈ M, then Xτ ∈ M. Indeed, if is an -sequence in X with dense span, then is an -sequence in X_{τ} with dense span. Of course, X_{σ} never belongs to M. On the other hand, it is well-known that L(X_{σ})=L(X_{τ}). Moreover, since σ(X, X′) ⊆ τ(X, X′), then any strongly continuous hereditarily hypercyclic operator semigroup on Xτ is also strongly continuous and hereditarily hypercyclic as an operator semigroup on X_{σ}. Thus in the case X_{τ} ∈ M, Theorem 1.3 implies that there is a strongly continuous hereditarily hypercyclic operator semigroup on X_{σ}. Unfortunately, the nature of the weak topology does not allow to make such a semigroup uniformly continuous.

It is worth noting that any separable *Fréchet* space admits an - sequence with dense span. It is also well-known [17] that any *Fréchet* space carries the Mackey topology and the topology on a *Fréchet* space X differs from the weak topology if and only if X is infinite dimensional and is non-isomorphic to ω. That is, any separable infinite dimensional *Fréchet* space non-isomorphic to ω belongs to M. Similarly, one can verify that an infinite dimensional inductive limit X of a sequence Xn for n ∈ of separable Banach spaces belongs to M provided there is n ∈ for which X_{n} is dense in X. Thus all the above mentioned existence theorems are particular cases of Theorem 1.3.

Grivaux [3] raised a question whether each separable infinite dimensional Banach space supports a hypercyclic non-mixing operator. Since the class M contains separable infinite dimensional Banach spaces, the following theorem provides an affirmative answer to this question.

**Theorem 1.6**. *Let X ∈ M. Then there exists T ∈ L(X) such that T is hypercyclic and non-mixing.*

The simplest separable infinite dimensional locally convex space space (and the only *Fréchet* space) outside M is ω. Curiously, the situation with ω is totally different. Hypercyclic operators on the complex space ω have been characterized by Herzog and Lemmert [18]. Namely, they proved that a continuous linear operator T on the complex *Fréchet* space ω is hypercyclic if and only if the point spectrum σ_{p}(T′) of T′ is empty. It also worth mentioning that Bés and Conejero [19] provided sufficient conditions for T ∈ L(ω) to have an infinite dimensional closed linear subspace, each non-zero vector of which is hypercyclic, and found common hypercyclic vectors for some families of hypercyclic operators on ω. See also the related work [20] by Petersson. The following theorem extends the result of Herzog and Lemmert and highlights the difference between ω and other *Fréchet* spaces.

**Theorem 1.7.** *Let T ∈ L(ω) be such that T′ has no non-trivial finite dimensional invariant subspaces and be a sequence of polynomials such that deg p _{l} → ∞ as l → ∞. Then the family {p_{l} (T) : l ∈ } is universal. Moreover, there is no strongly continuous supercyclic semigroup on ω.*

Note that in the case *=*, T′ has no non-trivial finite dimensional invariant subspaces if and only if σ_{p}(T′)=Φ. The first part of the above theorem implies that any hypercyclic operator on ω is mixing. We shall, in fact, verify the following more general statement.

**Theorem 1.8.** *Let X be a locally convex space carrying weak topology and T ∈ L(X). Then the following conditions are equivalent*

(1.8.1) *T′ has no non-trivial finite dimensional invariant subspaces;*

(1.8.2) *T is transitive;*

(1.8.3) *T is mixing;*

(1.8.4) *the semigroup is mixing, where is the multiplicative semigroup of non-zero polynomials with the norm |p|=deg p.*

**Remark 1.9.** Chan and Sanders [21] observed that on the space ()_{σ}, being the Hilbert space with the weak topology, there is a transitive non-hypercyclic operator. Theorem 1.8 provides a huge supply of such operators. For instance, the backward shift T on is mixing on ()σ (T′ has no non-trivial finite dimensional invariant subspaces) and T is clearly non-hypercyclic (each its orbit is bounded).

Theorems 1.3, 1.6 and 1.7 imply the following curious corollary.

**Corollary 1.10.** *Let X be a separable infinite dimensional Fréchet space. Then the following are equivalent*

(1.10.1) *there is a hypercyclic non-mixing operator* T ∈ L(X);

(1.10.2) *there is a mixing uniformly continuous semigroup on X ;*

(1.10.3) there is a supercyclic strongly continuous semigroup on X;

(1.10.4) *X is non-isomorphic to ω.*

Another simple space outside M is φ. Bonet and Peris [11] observed that there are no supercyclic operators on φ. On the other hand, Bonet, Frerick, Peris and Wengenroth [22] constructed a hypercyclic operator on the locally convex direct sum of countably many copies of the Banach space . The space X is clearly an LB_{s}-space, is complete and non-metrizable. It is also easy to see that (there are no -sequences in X with dense span). We find sufficient conditions of existence and of non-existence of a hypercyclic operator on a locally convex space. These conditions allow us to characterize the LBs-spaces, which admit a hypercyclic operator.

**Theorem 1.11.** *Let X be the inductive limit of a sequence of separable Banach spaces. Then the following conditions are equivalent:*

(1.11.1) *X admits no hypercyclic operator;*

(1.11.2) *X admits no cyclic operator with dense range;*

(1.11.3) *X is isomorphic to Y ×φ, where Y is the inductive limit of a sequence of separable*

*Banach spaces such that Y _{0} is dense in Y;*

(1.11.4) for any sufficiently large n, is finite dimensional and the set is infinite, where is the closure of X_{k} in X.

The proof is based upon the following result, which is of independent interest.

**Theorem 1.12.** *Let X be a topological vector space, which has no quotients isomorphic to φ. Then there is no cyclic operator with dense range on X ×φ.*

The following theorem provides another generalization of the mentioned result of Bonet, Frerick, Peris and Wengenroth.

**Theorem 1.13.** *Let be a sequence of separable Fréchet spaces. Then there is a hypercyclic operator on if and only if the set {n ∈ : Xn is infinite dimensional} is infinite.*

We derive the above theorem from the following result, concerning more general spaces.

**Theorem 1.14.** Let X_{n} ∈ M for each n ∈ and * Then there is a hypercyclic operator on X.*

The next issue, we discuss, are dual hypercyclic operators. Let X be a locally convex space. Recall that X_{β}′ is the dual space X′ endowed with the strong topology β(X′, X). It is worth noting that if X is a normed space, then the strong topology on X′ coincides with the standard norm topology. Salas [23] has constructed an example of a hypercyclic operator T on such that both T and T′ are hypercyclic. This result motivated Petersson [24] to introduce the following definition. We say that a continuous linear operator T on a locally convex space X is *dual hypercyclic* if both T and T′ are hypercyclic on X and X_{β}′ respectively. Using the construction of Salas, Petersson proved that any infinite dimensional Banach space X with a monotonic and symmetric Schauder basis and with separable dual admits a dual hypercyclic operator. He also raised the following questions. Does there exist a dual hypercyclic operator on any infinite dimensional Banach space with separable dual? Does there exist a non-normable *Fréchet* space that admits a dual hypercyclic operator? The first of these questions was recently answered affirmatively by Salas [25]. The following theorem provides a sufficient condition for existence of a dual hypercyclic operator on a locally convex space.

**Theorem 1.15.** *Let X be an infinite dimensional locally convex space admitting an -sequence with dense span. Assume also that there is an -sequence with dense span in X _{β}′ and at least one of the following conditions is satisfied:*

(1.15.1)* the topology of X coincides with σ(X, X′);*

(1.15.2) *the topology of X coincides with τ (X, X′);*

(1.15.3) *the set {fn : n ∈ } is uniformly equicontinuous.*

*Then X admits a dual hypercyclic operator.*

Since every separable *Fréchet* space admits an -sequence with dense span and every *Fréchet* space carries the Mackey topology, the above theorem implies the following corollary.

**Corollary 1.16.** *Let X be a separable infinite dimensional Fréchet space, such that there is an -sequence with dense span in X _{β}′. Then there exists a dual hypercyclic operator T ∈ L(X).*

If X is a Banach space, *X _{β}*′ is also a Banach space and therefore has an -sequence with dense span if and only if it is separable. Thus Corollary 1.16 implies the next corollary, which is the mentioned recent result of Salas.

**Corollary 1.17.** *Let X be an infinite dimensional Banach space with separable dual. Then there exists a dual hypercyclic operator T ∈ L(X).*

Corollary 1.16 also provides plenty of non-normable Fréchet spaces admitting a dual hypercyclic operator, thus answering the second of the above questions of Petersson. For instance, take the complex Fr´echet space X of entire functions on one variable with the topology of uniform convergence on compact sets. It is easy to verify that the sequence of functionals g_{n}(f)=(n!)^{−1f(n)}(0) is an -sequence with dense span in *X _{β}*′. Since X is also infinite dimensional and separable,Corollary 1.16 implies that X supports a dual hypercyclic operator.

The proofs of the above results are based upon the two main ingredients. One of them are sufficient conditions of mixing and the other is a criterion for a generic (in the Baire category sense) operator from a given class to be hypercyclic. Our sufficient conditions of mixing extend the result of Salas on hypercyclicity of perturbations of the identity by adding a backward weighted shift. Apart from providing us with tools, these extensions are of independent interest.

**Theorem 1.18.** *Let X be a topological vector space and T ∈ L(X) be such that the space*

(1.1)

*is dense in X. Then T is mixing. If additionally, X is Baire, separable and metrizable, then T is hereditarily hypercyclic.*

We shall see that the above theorem implies not only the mentioned result of Salas, but also is applicable in many other situations. For instance, we use the above theorem to prove the following results.

**Theorem 1.19.** Let X be a separable infinite dimensional Banach space and be the operator norm closure in L(X) of the set of finite rank nilpotent operators. Then the set of T ∈ for which T is supercyclic and I + T is hypercyclic is a dense G_{δ} subset of the complete metric space . If additionally X′ is separable, then the set of T ∈ for which T and T′ are supercyclic and I + T and I + T′are hypercyclic is a dense G_{δ} subset of .

Note that if a Banach space X has the approximation property [26], then the set from the above corollary is exactly the set of compact quasinilpotent operators (in the case = by quasinilpotency of T we mean quasinilpotency of the complexification of T or equivalently that ||T^{n}||^{1/n} → 0). Thus we have the following corollary.

**Corollary 1.20.** *Let X be a separable infinite dimensional Banach space with the approximation property and ⊂ L(X) be the set of compact quasinilpotent operators. Then the set of T ∈ for which T is supercyclic and I + T is hypercyclic is a dense G _{δ} subset of the complete metric space . If additionally X′ is separable, then the set of T ∈ for which T and T′ are supercyclic and I + T and I + T′ are hypercyclic is a dense G_{δ} subset of .*

**Theorem 1.21.** *Let X be a separable infinite dimensional Banach space and be the set of nuclear quasinilpotent operators endowed with the nuclear norm metric. Then the set of T ∈ for which T is supercyclic and I + T is hypercyclic is a dense G _{δ} subset of the complete metric space . If additionally X′ is separable, then the set of T ∈ for which T and T′ are supercyclic and I + T and I + T′ are hypercyclic is a dense G_{δ} subset of .*

Theorems 1.21 and 1.19 provide a large supply of dual hypercyclic operators T on any infinite dimensional Banach space with separable dual.

Godefroy and Shapiro [27] have introduced the notion of a generalized backward shift. Namely, a continuous linear operator T on a topological vector space X is called a *generalized backward shift* if its *generalized kernel*

is dense in X and ker T is one-dimensional. We introduce a more general concept. Namely, we say that T is an *extended backward shift* if

(2.1)

is dense in X. From the easy dimension argument [27] it follows that if T ∈ L(X) is a generalized backward shift, then dimker T^{n}=n and T (kerT^{n+1})=ker T^{n} for each n ∈ . Hence ker T^{n}=T^{n} (ker T^{2n}) and therefore ker T^{n}=T^{n} (X) ∩ ker T^{n} for any n ∈ . It follows that ker*T=ker^{†}T for a generalized backward shift. That is, any generalized backward shift is an extended backward shift.

We also consider the following analog of the concept of an extended backward shift for a k-tuple of operators. Let T_{1},...,T_{k} be continuous linear operators on a topological vector space X. We say that *T=(T _{1},...,T_{k}) ∈ L(X)^{k}* is a

(2.2)

It is easy to see that in the case of one operator (that is, k=1 and *T=T _{1} ∈ L(X)), κ (n,T) = T^{n} (kerT^{2n})=kerT^{n} ∩T^{n}(X)* and therefore the last definition is a generalization of the previous one. In order to study extended backward shifts we need to establish some properties of the backward shift on the finite dimensional space

**Backward shift on ^{2n}**

The following lemma is a modification of a lemma from reference [28].

**Lemma 2.1.** *For each n ∈ and z ∈ \{0}, the matrix is invertible.*

**Proof.** For each n, k ∈ consider the matrix First, we demonstrate that the determinants of M_{n,k} satisfy the recurrent formula

(2.3)

The equality (2.3) for n=2 is trivial. Suppose now that n ≥ 3. Subtracting the previous column from each column of M_{n,k} except the first one, we see that det M_{n,k}=det N_{n,k} where . Dividing the j-th row of N_{n,k} by j and multiplying the j-th column by a_{j}=(k + n − j + 1)(k + n − j) for 1 ≤ j ≤ n, we arrive at the matrix M_{n−1,k+2}. Hence

which proves (2.3). Since det M_{1,k}=1 for each k ∈ , from (2.3), it follows that det M_{n,k} ≠ 0 for any n, k ∈ . Let now B_{n} be the matrix obtained from A_{n,1} by putting the columns of A_{n,1} in the reverse order. Clearly det A_{n,1}=(−1)^{n−1} det B_{n}. On the other hand, multiplying the j-th column of B_{n} by (n − j + 1)! for 1 ≤ j ≤ n, we get the matrix M_{n,1}. Hence,

Since det M_{n,1} ≠ 0, we see that det A_{n,1} ≠ 0 and therefore An,1 is invertible. Finally, for any z ∈ consider the diagonal n × n matrix D_{n,z} with the entries (1, z,..., z^{n−1}) on the main diagonal. It is straightforward to verify that

A_{n,z}=zD_{n,z} A_{n,1}D_{n,z} for any z ∈ . (2.4)

Since A_{n,1} and D_{n,z} for z ≠ 0 are invertible, we see that A_{n,z} is invertible for each n ∈ and z ∈\ {0}.

**Lemma 2.2.** *Let n ∈ and e1,...,e2n be the canonical basis of ^{2n} and S ∈ L(^{2n}) be the backward shift defined by Se_{1}=0 and Se_{k}=e_{k−1} for 2 ≤ k ≤ 2n and P the linear projection on ^{2n} onto the subspace E=span {e_{1},...,e_{n}} along F=span {e_{n+1},...,e_{2n}}. Then for any z ∈\ {0} and u, v ∈ E, there exists a unique x^{z}=x^{z}(u, v) ∈ ^{2n} such that*

Px^{z}=u and Pe^{zS}x^{z}=v. (2.5)

*Moreover, for any bounded subset B of E and any ε > 0, there is c=c(ε, B) > 0 such that*

(2.6)

(2.7)

*In particular, xz (u, v) → u and e ^{zS} x^{z}(u, v) → v as |z| → ∞ uniformly for u, v from any bounded subset of E.*

**Proof.** Let u, v ∈ E and z ∈ \ {0}. For y ∈ ^{2n} we denote

One easily sees that (2.5) is equivalent to the vector equation

(2.8)

where is the matrix from Lemma 2.1 and is defined as

provided we set x_{j}=u_{j} for 1 ≤ j ≤ n. According to Lemma 2.1, the matrix is invertible for any and therefore (2.8) is uniquely solvable. Thus there exists a unique satisfying (2.5). It remains to verify the estimates (2.6) and (2.7). From (2.9) it follows that for any bounded subset B of E and any ε > 0, there is a=a(ε, B) such that

(2.10)

Recall that D_{n,z} is the diagonal n × n matrix with the entries (1, z,...,z^{n−1}) on the diagonal. Equalities (2.8) and (2.4) imply

where we use invertibility of A_{n,1} provided by Lemma 2.1. According to (2.10), the set is bounded in Hence the set is bounded in . From the last display we see that

Boundedness of Q implies now that (2.6) is satisfied with some c=c_{1}(ε, B). Finally, since

there exists c=c_{2}(ε, B) for which (2.7) is satisfied. Hence both (2.7) and (2.6) are satisfied with

c=max{ c_{1}(ε, B), c_{2}(ε, B)}.

The next corollary follows immediately from Lemma 2.2.

Corollary 2.3. Let and be as in Lemma 2.2. Then for any u, v ∈ E and any sequence satisfying there exists a sequence of elements of such that x_{j} → u and

Lemma 2.4. Let and be as in Lemma 2.2. Then for any bounded sequences of elements of E, there exists a sequence of elements of such that x_{j} − u_{j} → 0 and

**Proof.** It is easy to see that there is such that J has an upper triangular matrix, J is invertible and Indeed, S and e^{S} − I are similar since they are nilpotent of maximal rank 2n − 1. Moreover, since S and e^{S} − I are upper triangular, the similarity operator can be chosen upper triangular and therefore J (E) ⊆ E.

Let now x_{j}=J^{−1} x^{j} (Ju_{j}, Jv_{j}), where x^{z} (u, v) is defined in Lemma 2.2. Since the set

is bounded and is contained in E because J (E) ⊆ E, from Lemma 2.2 it follows that x^{j} (Ju_{j}, Jv_{j}) − Ju_{j} → 0 and e^{jS} x^{j} (Ju_{j}, Jv_{j}) − Jv_{j} → 0 as j → ∞. Multiplying by J^{−1}, we obtain x_{j} − u_{j} → 0 and J^{−1} e^{jS} Jx_{j}(I+S)^{j} x_{j} − v_{j} → 0 as j → ∞.

In order to construct multi-parameter **mixing semigroups**, we need the following multi-operator version of Corollary 2.3.

**Lemma 2.5.** Let for each j ∈ {1,..., k} let be the canonical basis in be the backward shift: S_{j} e^{j}_{1}=0 and 2 ≤ l ≤ 2n_{j}. Let also and for 1 ≤ j ≤ k,

where S_{j} sits in the j^{th} place. Finally, let be a sequence of elements of with Then for any u, v ∈ E, there exists a sequence of elements of X such that x_{m} → u and

**Proof.** If the statement of the lemma is false, then there are u, v ∈ E and a subsequence such that (u, v) does not belong to the closure of the set

Let be the one-point compactification of . Since is compact and metrizable, we can pick a convergent in subsequence Clearly the statement of the lemma remains false with {z_{m}} replaced by {z_{m}”}. That is, it suffices to consider the case when {z_{m}} converges in .

Thus without loss of generality, we can assume that {z_{m}} converges to w ∈ . Let C={j : w_{j}=∞}. Since the set C is non-empty. Without loss of generality, we may also assume that C={1,..., r} with 1 ≤ r ≤ k.

Denote by Σ the set of (u, v) ∈ X × X such that there exists a sequence of elements of X for which x_{m} → u and We have to demonstrate that E × E ⊆ Σ. Let uj ∈ Ej for 1 ≤ j ≤ k and By Corollary 2.3, for 1 j ≤ r, there exist sequences of elements of such that

Now we put and Consider the sequences and of elements of X defined by the formula According to the definition of x_{m} and y_{m} and the above display, x_{m} → 0 and y_{m} → u. Indeed, x_{m} → 0 because for any j, the sequence x_{j,m} is bounded and x_{1,m} → 0. Similarly, taking into account that (z_{m})_{j} → w_{j} for j > r, we see that . Hence (u, 0) ∈ Σ and (0, u) ∈ Σ. Thus

where On the other hand, it is easy to see that the linear span of the set ({0} × E_{0}) ∪ (E_{0} × {0}) is exactly E × E. Since Σ is a linear space, the above display implies that E × E ⊆ Σ.

For applications it is more convenient to reformulate the above lemma in the coordinate form.

**Corollary 2.6.** Let for each j ∈ {1,..., k}let N_{j}={1,..., 2n_{j}}, Q_{j} ={1,..., n_{j}}. Consider the sets M=N_{1} ×... × N_{k}, M_{0}=Q_{1} ×... × Q_{k} and let {em : m ∈ M} be the canonical basis of the finite dimensional vector space Let E=span {em : m ∈ M_{0}} and for 1 ≤ j ≤ k, T_{j} ∈ L(X) be the operator acting on the canonical basis in the following way: T_{j}e_{m}=0 if m_{j}=1, T_{j} e_{m}=e_{m′} if m_{j} > 1, where m_{l}′=m′_{l} if l ≠ j, m_{j}′=m_{j} − 1. Then for any sequence of elements of and any u, v ∈ E, there exists a sequence of elements of X such that x_{m} → u and

**The key lemmas**

**Lemma 2.7.** Let X be a Hausdorff topological vector space, and A=(A_{1},..., A_{k}) ∈ L(X)^{k} be such that A_{j} A_{l}=A_{l} A_{j} for any l, j ∈ {1,..., k}. Then for each x from the space κ(n, A) defined by (2.2), there exists a common finite dimensional invariant subspace for A_{1},..., A_{k} such that for any sequence of elements of there exist sequences , of elements of Y for which

(2.11)

where

**Proof. **Since x ∈ κ(n, T), there exists y ∈ X such that y = 0 for 1 ≤ j ≤ k. For each j ∈ {1,..., k} let Nj={1,..., 2nj} and Q_{j}={1,..., n_{j}}. Denote M=N_{1} ×... × N_{k}, M_{0}=Q_{1} ×... × Q_{k}. For any l ∈ M, let yand let Y=span {h_{l} : l∈ M}. Clearly Y is finite dimensional. It is also straightforward to verify that A_{j}h_{l}=0 if l_{j}=1 and A_{j} h_{l}=h_{l}′ if l_{j} > 1, where l′_{r}=l_{r} for r ≠ j, l′_{j}=l_{j} − 1, 1 ≤ j ≤ k_{0}. It follows that Y is invariant for A_{j} for 1 ≤ j ≤ k. Consider the linear operator defined on the canonical basis by the formulas Je_{l}=h_{l} for l ∈ M. Let also E=span {e_{l} : l ∈ M_{0}} and be the operators from Corollary 2.6. Taking into account the definition of Tj and the action of A_{j} on h_{l}, we see that A_{j}J=JT_{j} for 1 ≤ j ≤ k. Clearly n=(n_{1},..., n_{k}) ∈ M_{0} and therefore e_{n} ∈ E. Since , we also see that x=h_{n}. According to Corollary 2.6, there exist sequences of elements of such that and . Now for Then {x_{m}} and {y_{m}} are sequences of elements of Y. From the intertwining relations A_{j}J=JT_{j} and the fact that and Y are finite dimensional, it follows that , Thus (2.11) is satisfied.

Let X be a topological vector space and We write T ∈ ε(k, X) if for ∞ any the series converges in X and the map from is separately continuous, where and . The following proposition is an elementary exercise. We leave its proof for the reader.

**Proposition 2.8.** X be a locally convex space and T ∈ ε (k, X). Assume also that T_{j} T_{m} = T_{m} T_{j} for any m, j ∈ {1,..., k}. Then is a strongly continuous operator group. Moreover, if then the map is holomorphic for each x ∈ X.

**Remark 2.9.** It is worth noting that the semigroup property fails if the operators T_{j} are not pairwise commuting.

**Corollary 2.10.** Let X be a locally convex space, and A=(A_{1},..., A_{k}) ∈ ε (k, X) be such that A_{j} A_{l}=A_{l} A_{j} for any l, j ∈ {1,..., k}. Then for each x, y from the space ker^{†} (A) defined by (2.2) and any sequence of elements of there exist a sequence in X such that u_{m} → x and as m → ∞.

Proof. Fix a sequence of elements of Let Σ be the set of (x, y) ∈ X × X for which there exists a sequence in X such that u_{m} → x and as m → ∞. According to Lemma 2.7, κ(n, A) × {0} ⊆ Σ and {0} × κ(n, A) ⊆ Σ for any n=(n_{1},..., n_{k}) ∈ where the space κ(n, A) is defined by (2.2). On the other hand, from the definition of Σ it is clear that Σ is a linear subspace of X × X. Thus

Hence (x, y) ∈ Σ for any x, y ∈ ker† (A).

**Lemma 2.11.** Let X be a topological vector space,

There exist sequences of elements of X such that

(2.12)

Proof. If x=0, we can take u_{k}=v_{k}=0, so we may assume that x ≠ 0. Let n be the smallest positive integer for which A^{n}x=0. Since A^{m}x=0, we have n ≤ m. Hence x ∈ A^{m}(X) ⊆ A^{n}(X). Thus we can pick w ∈ X such that A^{n}w=x. Denote

Clearly Ah_{j}=h_{j}-1 for 2 ≤ j ≤ 2n, h_{n}=A^{n}h_{2n}=A^{n}w=x and Ah_{1}=A^{2n}h_{2n}=A^{n}x=0. By definition of n we have h_{1}=A^{2n−1}h_{2n}=A^{n−1}x ≠ 0. In particular Since the order of nilpotency of a nilpotent operator on Y cannot exceed the dimension of Y, we have dim Y ≥ 2n. On the other hand Y is the span of the 2n-elements set {h_{1},..., h_{2n}}. Hence {h_{1},..., h_{2n}} is a linear basis of Y. Thus there exists a unique linear isomorphism such that Je_{k}=h_{k} for 1 ≤ j ≤ 2n. Since Ah_{j}=h_{j−1} for 2 ≤ j ≤ 2n and Ah_{1}=0, we have A|_{y}=JSJ^{−1}, where is the backward shift operator from Lemma 2.2. Applying Lemma 2.4 with um=(0,..., 0, 1) and vm=(0,..., 0, 0), we find that there exists a sequence of vectors in such that g_{k} → e_{n} and (I + S)^{k}g_{k} → 0 as k → ∞. Applying Lemma 2.4 with u^{m}=(0,..., 0, 0) and v^{m}=(0,..., 0, z^{−m}), there exists also a sequence of vectors in such that f_{k} → 0 and z^{k} (I + S)^{k}f_{k} → e_{n} as k → ∞. Define now u_{k}=Jf_{k} and v_{k}=Jg_{k} for Since Y and are finite dimensional, we see that

Thus the sequences {u_{k}} and {v_{k}} satisfy the desired conditions.

The following corollary of Lemma 2.11 seems to be of independent interest.

Theorem 2.12. Let X be a topological vector space, T ∈ L(X) and Λ(T) be the set defined in (1.1). Then for any u, v ∈ Λ(T), there exists a sequence of elements of X such that x_{k} → x and Tk x_{k} → y. Pick

By Lemma 2.11, there exist sequences in X satisfying (2.12). Since I + A=z^{−1} T, (2.12) can be rewritten in the following way:

This shows that for any

On the other hand, the fact that Σ is a linear subspace of X × X, the definition of Λ(T) and the above display imply that Λ(T) × Λ(T) ⊆ Σ.

**Mixing semigroups and extended backward shifts**

We start by proving Proposition 1.1. The next observation is Proposition 1 in reference [28].

**Proposition G.** *Let X be a topological space and be a family of continuous maps from X to X such that T _{α}T_{β}=T_{β}T_{α} for any α, β ∈ A and T_{α}(X) is dense in X for any α ∈ A. Then the set of universal elements for F is either empty or dense in X.*

The following general theorem can be found in reference [28].

**Theorem U.** *Let X be a Baire topological space, Y a second countable topological space and {T _{a}: a ∈ A} a family of continuous maps from X into Y. Then the following assertions are equivalent:*

(U_{1}) *The set of universal elements for {T _{a} : a ∈ A} is dense in X;*

(U_{2}) *The set of universal elements for {T _{a} : a ∈ A} is a dense G_{δ} -subset of X;*

(U_{3}) *The set {(x, T _{a}x): x ∈ X, a ∈ A} is dense in X × Y.*

**Proof of Proposition 1.1.** Assume that {T_{t}}_{t∈A} is hereditarily hypercyclic. That is, is universal for any sequence of elements of A such that |tn| → ∞. Applying this property to the sequence t_{n}=nt with t ∈ A, |t| > 0, we see that T_{t} is hypercyclic. Since any hypercyclic operator has dense range [28], we get that T_{t}(X) is dense in X for any t ∈ A with |t| > 0. We proceed by reasoning ad absurdum. Assume that {T_{t}}_{t}∈_{A} is non-mixing. Then there exist nonempty open subsets U and V of X and a sequence of elements of A such that |t_{n}| → ∞ and for each Since each T_{tn} has dense range and T_{tn} commute with each other, Proposition G implies that the set W of universal elements of is either empty or dense in X. Since is universal, W is non-empty and therefore dense in X. Hence we can pick x ∈ W ∩ U. Since x is universal for , there is for which T_{tn} x ∈ V. Hence . This contradiction completes the proof of (1.1.1).

Next, assume that X is Baire separable and metrizable, {T_{t}}_{t}∈_{A} is mixing and is a sequence of elements of A such that |t_{n}| → ∞. From the definition of mixing it follows that for any non-empty open subsets U and V of X, for all sufficiently large Hence is dense in X × X. By Theorem U, is universal.

**Theorem 2.13.** Let X be a topological vector space, and A=(A_{1},..., A_{k}) ∈ ε (k, X) be a EBS_{k} -tuple. Then the strongly continuous group is mixing. If additionally X is Baire separable and metrizable, is hereditarily hypercyclic.

**Proof.** Assume that is non-mixing. Then we can find nonempty open subsets U and V of X and a sequence in such that |z_{m}| → ∞ as m → ∞ and for each Let Σ be the set of pairs (x, y) ∈ X × X for which there exists a sequence of elements of X such that x_{m} → x and . According to Corollary 2.10, ker^{†} (A) × ker^{†} (A) ⊆ Σ. Since A=(A_{1},..., A_{k}) is a EBS_{k} -tuple, ker^{†} (A) is dense in X and therefore Σ is dense in X × X. In particular, Σ meets U × V, which is not possible since for any This contradiction shows that is mixing. If X is Baire separable and metrizable, Proposition 1.1 implies that is hereditarily hypercyclic.

It is easy to see that if X is a Banach space and Moreover, each operator group of the shape is uniformly continuous. Hence, we get the following corollary of Theorem 2.13.

**Corollary 2.14.** Let X be a separable Banach space and (A_{1},..., A_{k}) ∈ L(X)^{k} be a EBS_{k} -tuple. Then is a hereditarily hypercyclic uniformly continuous group.

**Lemma 3.1.** Let Y_{0} and Y_{1} be closed linear subspaces of a locally convex space Y such that Y_{0} ⊂ Y_{1} and the topology of Y_{1}/Y_{0} is not weak. Then there is a sequence in Y′such that

(3.1.1)

(3.1.2) for each

(3.1.3) is uniformly equicontinuous.

**Proof.** Since the topology of Y_{1}/Y_{0} is not weak, there exists a continuous seminorm on Y1/Y0 such that the closed linear space ker =^{-1} (0) has infinite codimension in Y_{1}/Y_{0}. Clearly the seminorm p on Y_{1} defined by the formula is also continuous and ker p has infinite codimension and contains Y_{0}. In particular the space Y_{p}=Y_{1}/ker p endowed with the norm is an infinite dimensional normed space. Hence we can choose sequences in Y_{1} and in Y′_{p} such that ||gn|| ≤ 1 for each and g_{n}(y_{k} + ker p)=δ_{n,k} for n, where δ_{n,k} is the Kronecker δ (every infinite dimensional normed space admits a biorthogonal sequence). Now let the functionals be defined by the formula h_{n}(y)=g_{n}(y_{k} + ker p). The above properties of the functionals g_{n} can be rewritten in terms of h_{n} in the following way for any n,

Since any continuous seminorm on a subspace of a locally convex space extends to a continuous seminorm on the entire space [17,29], we can find a continuous seminorm q on Y such that q|Y_{1} = p .

Applying the Hahn–Banach theorem, we can find such that

From the last two displays we have f_{n}(y_{k})=δ_{n,k}, which implies (3.1.1) since y_{k} ∈ Y_{1}. From the inequality in the above display it follows that each |f_{n}| is bounded by 1 on the unit ball W of the seminorm q. Since W is a neighborhood of zero in Y, condition (3.1.3) is satisfied. Since Y_{0} ⊆ ker p ⊆ ker q, from the inequality it follows that each f_{n} vanishes on Y_{0}. That is, (3.1.2) is satisfied.

Applying Lemma 3.1 with Y_{0}=0 and Y_{1}=Y, we obtain the following corollary.

**Corollary 3.2.** *Let Y be a locally convex space, whose topology is not weak. Then there exists a linearly independent sequence in Y′ such that is uniformly equicontinuous and *

Recall that a subset D of a locally convex space X is called a disk if D is bounded, convex and balanced (=is stable under multiplication by any The symbol X_{D} stands for the space span (D) endowed with the norm being the Minkowskii functional of the set D. Boundedness of D implies that the topology of X_{D} is stronger than the one inherited from X. A disk D in X is called a Banach disk if the normed space X_{D} is complete. It is well-known that a sequentially complete disk is a Banach disk, see, for instance, [29]. In particular, a compact or a sequentially compact disk is a Banach disk. We say that D is a Banach s-disk in X if D is a Banach disk and the Banach space X_{D} is separable.

**Lemma 3.3.** Let be an sequence in a locally convex space X. Then the set

is a compact and metrizable disk. Moreover, K is a Banach s-disk and E=span is dense in the Banach space X_{K}.

**Proof. **Let be endowed with the coordinatewise convergence topology. It is easy to see that Q is a metrizable compact topological space as a closed subspace of

Obviously, the map is onto. It is also easy to see that φ is continuous. Indeed, let p be a continuous seminorm on X, a ∈ Q and ε > 0. Since x_{n} → 0, there is such that p(x_{n}) ≤ ε for n > m. Let δ=ε(1 + p(x_{0}) +... + p(x_{m}))^{−1} and W={b ∈ Q : |a_{j} − b_{j} | < δ for 0 ≤ j ≤ m}. Then W is a neighborhood of a in Q and for each b ∈ W we have

Taking into account that p(x_{n}) < ε for n > m and |a_{n} − b_{n}| < δ for n ≤ m, we obtain

Using the definition of δ and inequalities ||a||_{1} ≤ 1, ||b||_{1} ≤ 1, we see that p(φ(b) − φ(a)) ≤ 3ε. Since a, p and ε are arbitrary, the map φ is continuous. Hence K is compact and metrizable as a continuous image of a compact metrizable space. Obviously K is convex and balanced. Hence K is a Banach disk (any compact disk is a Banach disk). Let us show that E is dense in X_{K}. Take u ∈ X_{K}. Then there is a ∈ such that

Let || . || be the norm of the Banach space X_{K}. Then for any

Hence E is dense in X_{K} and therefore X_{K} is separable and K is a Banach s-disk.

**Lemma 3.4.** Let X be a separable metrizable topological vector space and be a linearly independent sequence in X. Then there exist sequences in X and in such that span is dense in X, g_{n}(x_{k})=0 for n ≠ k and g_{n}(x_{k}) ≠ 0 for where

**Proof.** Let be a base of topology of X. First, we construct inductively sequences

in and in X such that for any

(b1) y_{k} ∈ U_{k};

(b2) g_{k} (y_{k}) ≠ 0, where

(b3) g_{k} (y_{k})=0 if m < k.

Let g_{0}=f_{0}. Since f_{0} ≠ 0, there is y_{0} ∈ U_{0} such that f_{0}(y_{0})=g_{0}(y_{0}) ≠ 0. This provides us with the base of induction. Assume now that n ∈ and y_{k}, α_{k,j} with j < k < n satisfying (b1– b3) are already constructed. According to (b2) and (b3), we can find α_{n},0,..., α_{n,n−1} ∈ such that g_{n}(y_{m})=0 for m < n, where Next since f_{j} are linearly independent, g_{n} ≠ 0 and therefore there is yn ∈ Un such that g_{n}(y_{n}) ≠ 0. This concludes the description of the inductive procedure of constructing sequences satisfying (b1–b3) for each

Using (b2) and (b3), one can easily demonstrate that there is a sequence in such that g_{n}(x_{k})=0 for k ≠ m, . From (b2) and (b3) it also follows that g_{n}(x_{n}) 0 for each It remains to notice that according to (b1), is dense in X. Since ⊆ span =span we see that span is dense in X.

**Lemma 3.5.** Let Then there exist sequences and in X and X′ rspectively, such that

(3.5.1) is an -sequence in X;

3.5.2) the space E=span is dense in X;

(3.5.3) f_{k} (x_{n})=0 if k ≠ n and f_{k} (x_{k}) ≠ 0 for each

(3.5.4) is uniformly equicontinuous.

*Moreover, {f _{k}} can be chosen from the linear span of any linearly independent uniformly equicontinuous sequence in X′.*

**Proof.** According to Lemma 3.3, there exists a Banach s-disk K in X such that X_{K} is dense in X. By Corollary 3.2, there is a linearly independent sequence in X′ such that is uniformly equicontinuous. Since X_{K} is dense in X, the functionals on X_{K} are linearly independent. Applying Lemma 3.4 to the sequence , we find that there exist sequences in X_{K} and in such that E=span is dense in X_{K}, h_{n}(y_{k})=0 for n ≠k and h_{n}(y_{n}) ≠ 0 for where Let q be the norm of the Banach space X_{K}. Consider f_{n}=c_{n}h_{n}, where . Since is uniformly equicontinuous and we immediately see that is uniformly equicontinuous. Next, let Since x_{n} converges to 0 in the Banach space -sequence in X_{K}. Since the topology of X_{K} is stronger than the one inherited from X, -sequence in X. Since span{ is dense in X_{K}, it is also dense in X_{K} in the topology inherited from X. Since X_{K} is dense in X, span is dense in X. Finally since f_{n}(x_{k})=c_{n}b_{k} h_{n}(y_{k}), we see that f_{n}(x_{k})=0 if n ≠ k and f_{n}(x_{n}) ≠ 0 for any Thus conditions (3.5.1–3.5.4) are satisfied.

**Lemma 3.6.** *Let X be a locally convex space, whose topology is not weak and Y be a locally convex space admitting an -sequence with dense span. Then there is T ∈ L(Y, X) such that T (X) is dense in Y.*

**Proof.** Let be an -sequence in Y with dense span. By Corollary 3.2, there is a uniformly equicontinuous sequence in X′ such that ,. Consider the linear operator T : X → Y defined by the formula .Since {f_{n}} is uniformly equicontinuous, there is a continuous seminorm p on X such that |f_{n}(x)| ≤ p(x) for any x ∈ X and Since {y_{n}} is an -sequence, Lemma 3.3 implies that the closed convex balanced hull Q of is compact and metrizable. Let q be the Minkowskii functional of Q (=the norm on YQ). It is easy to see that q(T x) ≤ p(x) for each x ∈ X. Hence T is continuous as an operator from X to the Banach space YQ. Since the latter carries the topology stronger than the one inherited from Y, T ∈ L(X, Y). Next, the inclusion , implies that T (X) contains the linear span of , which is dense in Y. Thus T has dense range.

**Lemma 3.7.** Let X be an infinite dimensional locally convex space and assume that there exist -sequences with dense span in both X and X′_{β}. Then there exist sequences and in X and X′ respectively, such that.

(3.7.1) is an -sequence in X and is an -sequence in X′_{β};

(3.7.2) E=span is dense in X and F=span is dense in X′_{β};

(3.7.3) f_{k}(x_{n})=0 if k ≠ n and f_{k} (x_{k}) ≠ 0 for each

If the original -sequence in X′_{β} is uniformly equicontinuous, than we can also ensure that (3.7.4) is uniformly equicontinuous.

**Proof.** By Lemma 3.3, there exists a Banach s-disk K in X such that XK is dense in X. Fix an -sequence in X′_{β} with dense span. Since any sequence of elements of a linear space with infinite dimensional span has a linearly independent subsequence with the same span, we, passing to a subsequence, if necessary, can assume that gn are linearly independent. Since X_{K} is dense in X, the functionals on X_{K} are linearly independent. Let D be the closed absolutely convex hull of the set in X′_{β}By Lemma 3.3, D is a Banach disk in X′_{β} Applying Lemma 3.4 to the sequence , we find that there exist sequences in X_{K} and in such that E=span is dense in X_{K}, h_{n}(y_{k})=0 for n ≠ k and h_{n}(y_{n}) ≠ 0 for where Let q be the norm of the Banach space XK and p be the norm of the Banach space X′_{D}. Consider f_{n}=c_{n}h_{n}, where . Since p(g_{n}) ≤ 1 for each and , we immediately see that for any Hence is an -sequence in the Banach space X′ _{D}. Since the topology of X′ _{D} is stronger than the one inherited from X′_{β}, is an -sequence in X′_{β}. Similarly, is uniformly equicontinuous if is. Since the sequences have the same spans, the span of is dense in X′_{β}. Next, let x_{n}=b_{n}y_{n}, where b_{n}=2^{−n}q(x_{n})^{−1}. Since x_{n} converges to 0 in the Banach space X_{K}, is an -sequence in X_{K}. Since the topology of X_{K} is stronger than the one inherited from X, is an - sequence in X. Since span =span is dense in X_{K}, it is also dense in X_{K} in the topology inherited from X. Since X_{K} is dense in X, span is dense in X. Finally since f_{n}(x_{k})=c_{n}b_{k}h_{n}(y_{k}), we see that fn(x_{k})=0 if n ≠ k and f_{n}(x_{n}) ≠ 0 for any Thus conditions (3.7.1–3.5.3) (and (3.5.4) if is uniformly equicontinuous) are satisfied.

**Lemma 3.8.** The class is stable under finite or countable products. Moreover,

**Proof. **It is clear that if the topology of one of the locally convex spaces Xj is not weak, then so is the topology of their product Π_{X j} . Thus the only thing we have to worry about is the existence of an -sequence with dense span. Let J be a finite or countable infinite set, for each j ∈ J and . Let for any j ∈ J, be an -sequence in Xj with dense span. For (j, n) ∈ let u_{j,n} ∈ X be such that j^{th} component of u_{j,n} is x_{j,n}, while all other components are 0. Clearly {u_{j,n} : j ∈ J,} is a countable subset of X. It is easy to see that enumerating this subset by elements of , we get an -sequence in X with dense span. The proof of the second part of the lemma is even easier: one have just to add one vector (0, 1) to an -sequence with dense span in X=X × {0}, to obtain an -sequence with dense span in X × .

Let By Lemma 3.5, there exist sequences and in X and X′ respectively satisfying (3.5.1–3.5.4). Uniform equicontinuity of {f_{n}} is equivalent to the existence of a continuous seminorm p on X such that each |f_{n}| is bounded by 1 on the unit ball {x ∈ X : p(x) ≤ 1} of p. Since {x_{n}} is an -sequence in X, Lemma 3.3, absolutely convex closed hull K of {x_{n} : } is compact and is a Banach disk in X. Let q be the norm of the Banach space X_{K}. Then q(x_{n}) ≤ 1 for each Let c=sup{p(x) : x ∈ K}. Compactness of K implies that c is finite. Clearly c > 0 and p(x) ≤ cq(x) for any x ∈ Y.

**Lemma 4.1.** Let α, β : be any maps and Then the formula

(4.1)

defines a linear operator on X. Moreover, the series converges in X for any x ∈ X anid and

for each x ∈ X and (4.2)

where ||a|| is the -norm of a.

**Proof. **Condition (3.5.4) implies that the sequence is bounded for any x ∈ X. Since {x_{n}} is an -sequence and a ∈ , we see that the series in (4.1) converges for any x ∈ X and therefore defines a linear operator on X. Moreover, if p(x) ≤ 1, then |f_{k} (x)| ≤ 1 for each and since q(x_{m}) ≤ 1 for formula (4.1) implies that q(Tx) ≤ ||a||. Hence q(Tx)≤ ||a||p(x) for each x ∈ X. Then q(T^{2}x) ≤ ||a|| p(Tx) ≤ c||a||q(Tx) ≤ c||a||^{2}p(x). Iterating this argument, we see that

q(T^{n}x) ≤ c^{n−1}||a||^{n} p(x) for each

It follows that for any x ∈ X, the series is absolutely convergent in the Banach space X_{K} (and therefore in X) for any . Naturally we denote its sum as e^{zT} x. Moreover, using the above display, we obtain

Thus (4.2) is satisfied.

Now fix a bijection By symbol e_{j} we denote the element of defined by , where is the Kronecker delta. For , we also write . For each let

According to (3.5.1), (3.5.3) and (3.5.4), {ε_{m}} is a bounded sequence positive numbers. Next, we pick any sequence of positive numbers such that

for any (4.3)

and consider the operators defined by the formula

From the estimates (4.3) it follows that the series defining A_{j} can be written as

with

Clearly A_{j} have shape (4.1) with .By Lemma 4.1, A_{j} are linear operators on X satisfying for any . Hence A_{j} are continuous as operators from X to the Banach space X_{K}. Since X_{K} carries a stronger topology than the one inherited from . Next, using the definition of A_{j , }it is easy to verify that for any and Indeed, for any , there is a unique such that If either , from the definition of A_{j }and A_{i }it follows that . If from the same definition we obtain .From (3.5.2) it follows now that the operators are pairwise commuting. Next, for , the operator has shape (4.1) with , where is the -norm of z. By Lemma 4.1, the series converges pointwisely to a linear operator and

for any (4.4)

Exactly as for the operators A_{j}, we see that for any From (4.4), if uniformly on By Proposition 2.8, is a uniformly continuous group and the map is holomorphic for any provided The proof will be complete if we show that the group is hereditarily hypercyclic. To this end consider the restrictions B_{j }of A_{j }to X_{K} as bounded linear operators on the Banach space X_{K}. Then B_{j } commute with each other as the restrictions of commuting operators. Let us show that the operator group on X_{K }is hereditarily hypercyclic. By Corollary 2.14, it suffices to verify that is a EBSk -tuple. We already know that B_{j} are pairwise commuting. From the definition of B_{j} it is easy to see that ker contains .Using this fact it is easy to see that for any the set defined by (2.2) contains

Hence the space defined in (2.2) contains . By Lemma 3.3 is dense in X_{K}. Hence B is an EBS_{K} -tuple. Thus according to Corollary 2.14, is hereditarily hypercyclic. Since X_{K} is dense in X and carries a topology stronger than the one inherited from X, is hereditarily hypercyclic. By Proposition 1.1,

**Theorem 1.18: proof and applications**

Proof of Theorem 1.18. Assume that T is non-mixing. Then we can choose non-empty open subsets *U* and *V *of *X* and a strictly increasing sequence of positive integers such that for any Let Σ be the set of for which there exists a sequence of elements of X such that According to Theorem 2.12, Σ contains Since Λ(T) is dense in X, we see that Σ is dense in X × X. Hence Σ intersects U × V, which is impossible since for any This contradiction shows that T is mixing. If X is Baire separable and metrizable, then by Proposition 1.1, T is hereditarily hypercyclic.

**Extensions of the Salas theorem**

It is easy to see that ker for any linear operator T. Thus Theorem 1.18 implies the following corollary.

**Corollary 5.1. **Let T be an extended backward shift on a topological vector space X. Then I + T is mixing. If additionally X is Baire, separable and metrizable, then I + T is hereditarily hypercyclic.

Recall that a backward weighted shift on is the operator T acting on the canonical basis as follows: where is a bounded sequence of non-zero numbers in *K*. Clearly any backward weighted shift is a generalized backward shift and therefore is an extended backward shift. Hence Corollary 5.1 contains the Salas theorem on hypercyclicity of the operators *I + T *with T being a backward weighted shift as a particular case. It is also easy to see that if *X* is a topological vector space and is surjective, then .Thus we obtain the following corollary.

**Corollary 5.2**. *Let X be a topological vector space and be such that T (X)=X and ker* T is dense in X. Then I + T is mixing. If additionally X is Baire, separable and metrizable, then I + T is hereditarily hypercyclic.*

We can further generalize Corollary 5.1 by means of the following observation.

** Lemma 5.3. **Let X be a topological vector space, k ∈ N, T ∈ L(X) and {a

*Proof. *In order to prove the inclusion , it is enough to show that for any n ∈ N Thus pick n ∈ N and let .Then and there is y ∈ X such that .For denote .It is easy to see that , for 2 ≤ j ≤ 2kn and . Consider the backward shift B on the space , which acts on the canonical basic vectors by the same rule: Be_{1}=0 and for 2 ≤ j ≤ 2kn. Let Y =span{hj : 1 ≤ j ≤ 2kn} and consider the surjective linear operator defined by for.It is easy to see that Y is T-invariant and ,where the restriction of T to Y. Since B^{2kn}=0, we see that and therefore the restriction to Y is given by the formula , where P is the polynomial defined by .Let E =span{e_{j} : 1 ≤ j ≤ k_{n}}. Considering the matrix of the operator p(B), it is easy to see that .Using the intertwining relation , we see that .Since is the restriction to Y of the operator T^{k} +S, we see that

The following result is an immediate consequence of Lemma 5.3 and Corollary 5.1.

**Corollary 5.4** Let X be a topological vector space, and be a sequence in K such that the series of operators converges pointwise to a continuous linear operator S on X. Assume also that is an extended backward shift. Then the operator I++S is mixing. If additionally X is Baire separable and metrizable, then I++S is hereditarily hypercyclic.

It is straightforward to verify that any power of a generalized weighted shift is an extended backward shift. Thus the above corollary implies the next observation

**Corollary 5.5** *Let X be a topological vector space, T ∈ L(X) be a generalized backward shift and be a sequence in K such that , there is n ∈ N for which and the series converges pointwise to a continuous linear operator S on X. Then S is mixing. If additionally X is Baire separable and metrizable, then S is hereditarily hypercyclic.*

**An extension of the Hilden–Wallen theorem **

**Lemma 5.6 **Let X be a Baire topological space, Y be a second countable topological space and be a sequence of continuous maps from X to Y. Let alsoΣ be the set of for which there exists a sequence of elements of X such that and as . If Σ is dense in X×Y, then is hereditarily universal.

**Proof. **The density of Σ in X×Y implies that for any infinite set ,condition (U3) from Theorem U is satisfied for the family and therefore this family is universal.

Hilden and Wallen [30] demonstrated that any backward weighted shift on is supercyclic. Many particular cases of the following proposition are known, see for instance [31]. We include it here in its full generality for the sake of completeness.

**Proposition 5.7** Let X be a Baire separable metrizable topological vector space *T ∈ L(X)*. Suppose also that T has dense range and dense generalized kernel. Then T is hereditarily supercyclic.

**Proof.** Since T has dense range, we have that T^{k}(X) is dense in X for each . Thus for any x ∈X there exists a sequence in X such that as .Fix a dense countable set .Since X is metrizable and B is countable, we can choose a sequence of positive numbers such that asfor each.

Let now for eachand Σ be the set of for which there exists a sequence of elements of X such that and .For any ,let .Then and for each . Hence . On the other hand, for any ,we have for all sufficiently large k and therefore . Considering the constant sequence ,we see that .

Finally, observe that Σ is a linear subspace of X×X. Hence .Since both B and ker *T are dense in X, we see that Σ is dense in X× X. By Lemma 5.6, for each infinite set.the family is universal. Hence T is hereditarily supercyclic.

**Remarks on Theorem 1.18**

Theorem 1.18 is reminiscent of the following criterion of hypercyclicity of Bayart and Grivaux [32] in terms of the unimodular point spectrum.

**Theorem BG ***Let X be a complex separable infinite dimensional Banach space, T ∈ L(X) and assume that there exists a continuous Borel probability measure μ on the unit circle T such that for each Borel set with μ(A)=1, the space*

*is dense in X. Then T is hypercyclic.*

It is worth noting that in the case of operators on Banach spaces, neither Theorem 1.18 implies the result of Bayart and Grivaux, nor their result implies Theorem 1.18, see Examples 5.9, 5.10 and 5.12 below. Theorem 1.18 is also strictly stronger than Proposition 2.2 in the article [33] by Herrero and Wang, which is a key tool in the proof of the main result in reference [33] that any element of the operator norm closure of the set of hypercyclic operators on *l*_{2} is a compact perturbation of a hypercyclic operator. Namely, they assume that the span taken in (1) is dense taking only into account the z’s for which T -zI has closed range, and this allows them to use the Kitai criterion. We would like also to mention the following fact.

**Proposition 5.8 ***Let X be a locally convex space and Y be a closed linear subspace of X such that X admits an l _{1}-sequence with dense span and the topology of X/Y is not weak. Then there is a hereditarily hypercyclic operator T ∈ L(X) such that Ty=y for any y ∈ Y.*

**Proof. **Since the topology of X/Y is not weak, Lemma 3.1 implies that there is a linearly independent uniformly equicontinuous sequence such that Y ⊆ ker g_{n} for each n ∈ Z_{+} .By Lemma 3.5, we can find sequence and in X and X′ respectively, such that conditions (3.5.1–3.5.4) are satisfied and each* f _{k}* belongs to span . Hence for any Uniform equicontinuity of {

defines a continuous linear operator on X, which also acts continuously on the Banach space X_{k}, where K is the Banach disk being the closed convex balanced hull of . By Lemma 3.3, X_{k} is separable and is dense in X_{k} .It is also straightforward to verify that contains E, where is the restriction of T to X_{k}. Hence T_{k} is an extended backward shift on the separable Banach space X_{k}. By Corollary 5.1, I+T_{k} is hereditarily hypercyclic. Since X_{k} is dense in X and carries stronger topology,is hereditarily hypercyclic. Since for each , from the definition of T it follows that for each y ∈ Y. Thus I + T satisfies all required conditions.

Grivaux [34] proved that if Y is a closed linear subspace of a separable Banach space X such that X/Y is infinite dimensional, then there exists a hypercyclic T ∈ L(X) such that Ty=y for any y ∈ Y. This result is an immediate corollary of Proposition 5.8. If we apply Proposition 5.8 in the case when X is a separable Fréchet space, we obtain that whenever Y is a closed linear subspace of X such that X/Y is infinite dimensional and non-isomorphic to ω, there is a hereditarily hypercyclic T ∈ L(X) such that Ty=y for any y ∈ Y. It is also worth noting that any non-normable Fréchet space has a quotient isomorphic to ω, [29]. If X is a non-normable separable Fréchet space non-isomorphic to ω and Y is a closed linear subspace of X such that X/Y is isomorphic to ω, then there is no hypercyclic operator T ∈ L(X) such that Ty=y for each y ∈ Y. Indeed, assume that such a T does exist. Then , is a continuous linear operator with dense range from X/Y to X. Since X/Y is isomorphic to ω and ω carries the minimal locally convex topology, S is onto and therefore X is isomorphic to ω, which is a contradiction.

**Example 5.9 ***Let be the complex Sobolev space which consists of the distributions f on T such that where the denotes the n ^{th} Fourier coefficient of f. Then the operator T acting on as satisfies the conditions of Theorem BG (and therefore is hypercyclic) and does not satisfy the conditions of Theorem 1.18.*

**Proof.** For each is the onedimensional space spanned by the Dirac δ -function δ_{z}, which does not belong to the range of .Thus and T does not satisfy conditions of Theorem 1.18. On the other hand, is dense in for any set which is dense in T. Since any subset of T of full Lebesgue measure is dense, the conditions Theorem BG are satisfied with μ being the normalized Lebesgue measure.

**Example 5.10 **If is the complex Sobolev space which consists of the distributions f on T such that ,then the operator acting on satisfies the conditions of both Theorem BG and Theorem 1.18.

**Proof. **For each and therefore the dense linear span of the set is contained in .Hence T satisfies the conditions of Theorem 1.18. As in the above example, is dense in for any set which is dense in T and therefore T satisfies conditions of Theorem BG.

**Example 5.11** If T is a quasinilpotent generalized backward shift on a separable complex Banach space, then *I+T* satisfies the conditions of Theorem 1.18 and does not satisfy the conditions of Theorem BG.

**Proof.** Since conditions of Theorem BG are not satisfied for the operator *I+T.* On the other hand, is dense and therefore I+T satisfies conditions of Theorem 1.18.

Examples of chaotic operators on a complex Hilbert space *H* which are not mixing are constructed in [35]. For such an operator, the linear span of the union of satisfying for some is dense in H. This shows that the assumption of Theorem 1.18 cannot be relaxed into a weaker assumption like density of the linear span of the union of . Note also that the class of operators T for which is dense is closed under finite direct sums. Moreover, this class for operators acting on Banach spaces is closed under infinite c_{0}-sums and *l _{p}*-sums for . In particular,

Now we describe another class of operators to which Theorem 1.18 applies. Let and be a Borel measurable map. Consider the integral operator defined by

It is straightforward to verify that is a compact linear operator on

**Example 5.12** Assume thatΨ is continuous, strictly increasing and that almost everywhere on [0.1]. Then acting on is hereditarily supercyclic and I+T is mixing.

**Proof.** Consider the sequence Clearly is strictly decreasing and tends to zero as .One can easily verify that

Using this equality it is straightforward to check that is dense in for each and that is dense in the entire space. Thus is dense in .That is, T is an extended backward shift. It remains to apply Proposition 5.7 and Corollary 5.1.

**Universality of generic families**

**Theorem 6.1** *Let X be a separable metrizable Baire topological space, Ω be a Baire topological space, A be a set, and for each let Ψ _{a} be a map from Ω into the set C(X, X) of continuous maps from X to X such that *

(6.1)

(6.2)

Then

**Proof. **Let be a sequence of non-empty open subsets of *X*, which form a basis of the topology of *X*. By Theorem *U*, is universal if and only if for each there exists for which .Thus we have

According to (18), the sets are open in* Ω.* Hence, the above display implies that *U* is a G_{δ} -subset of Ω. It remains to show that U is dense in Ω. Let

Clearly U is the projection of U_{0} onto Ω. On the other hand, U_{0} is the set of universal elements of the family .Since the product of two Baire spaces, one of which is second countable, is Baire [36], Ω × X is Baire. Applying Theorem U, we see that U_{0} is dense in Ω × X. Since the projection onto *Ω *of a dense subset of *Ω × X* is dense, we get that U is dense in *Ω.*

We apply the above general result to two types of universality: hypercyclicity and supercyclicity.

**Theorem 6.2** *Let X be a separable metrizable Baire topological vector space, Ω be a Baire topological space and be a map from Ω intoL(X)such that*

(6.3)

Then the following conditions are equivalent:

(6.4)

(6.5)

Proof. Let andfforbe defined as .Applying Theorem 6.1, we see that (6.4) implies (6.5). Since the set of hypercyclic vectors of any hypercyclic operator is dense, (6.5) implies (6.4).

**Theorem 6.3** *Let X be a separable metrizable Baire topological vector space, Ω be a Baire topological space and be a map from Ω to L(X) such that (6.3) is satisfied. Then the following conditions are equivalent:*

(6.6)

(6.7)

**Proof**. Let and for be defined as .Applying Theorem 6.1, we see that (6.6) implies (6.7). Since the set of supercyclic vectors of any supercyclic operator is dense, (6.7) implies (6.6).

**Corollary 6.4 ***Let X be a separable Baire metrizable topological vector space, Ω be a Baire topological space and be a map from Ω to L(X) satisfying (6.3). Suppose also that for any non-empty open subset W of Ω and any nonempty open subsets U, V of X, there exist , and such that ,where is defined in (1.1). Then the set of α ∈ Ω for which Tα is hypercyclic is a dense G -subset of Ω.*

**Proof.**Let be such that .By Theorem 2.12 (x, y) is in the closure of the set .Hence is in the closure of the set By the assumptions of the corollary, the last set is dense in Ω × X × X. It remains to apply Theorem 6.2.

The following supercyclicity analog of the above corollary turns out to be much easier.

**Proposition 6.5*** Let X be a Baire separable metrizable topological vector space, Ω be a Baire topological space and be a map from Ω to L(X) satisfying (6.3). Suppose also that for any non-empty open subset W of Ω and any nonempty open subsets U and V of X, there exist , and n ∈ N such that and . Then the set of for which T _{α }is supercyclic is a dense G_{δ} -subset of Ω.*

**Proof.** Let be such that there exists *n ∈ N *for which and .Let be such that For each consider .Then as .Moreover, for any .Thus for all sufficiently large m, .Since W, U and V were arbitrary, the set is dense in Ω × X × X. It remains to apply Theorem 6.3.

The obvious inclusion and the fact that a map satisfies (6.3) if and only if the map satisfies (6.3) imply the following corollary of Corollary 6.4 and Proposition 6.5.

**Corollary 6.6 ***Let X be a Baire separable metrizable topological vector space, Ω be a Baire topological space and be a map from Ω to L(X) satisfying (6.3). Suppose also that for any non-empty open subset W of Ω and any nonempty open subsets U and V of X, there exist , x ∈ U and y ∈ V such that .Then the set of for which _{ }is supercyclic and is hypercyclic is a dense G_{δ}-subset of Ω.*

Now we apply the above general results. In particular, we shall prove Theorems 1.19 and 1.21. For a while we shall assume that X and Y are two infinite dimensional Banach spaces, * *is a continuous bilinear form separating points of X and of Y. That is, for each non-zero x ∈ X, there is y ∈ Y satisfying * *and for each non-zero y ∈ Y, there is x ∈ X such that *. *In particular,b is a dual pairing between X and Y. Recall that the injective norm on the tensor product is defined by the formula

The completion of * *with respect to this norm is called the injective tensor product of X and is Y and denoted .It is again a Banach space. For each we consider the linear operators and defined by the formulae

(6.8)

Clearly and are bounded linear operators from endowed with the injective norm into the Banach spaces L(X) and L(Y) respectively. Hence they admit unique continuous extensions to , which we again denote by and. Note that is the dual operator of with respect to the dual pairing b:

(6.9)

Let now *M* be a linear subspace ofwhich we suppose to be endowed with its own *F*-space topology stronger than the one inherited from.Suppose also that X ⊗Y is a dense linear subspace of *M* and let be the closure in *M* of the set .Remark that (6.9) implies that if and only if . Thus nilpotency of is equivalent to nilpotency of and this implies that coincides with the closure in *M* of the set

**Proposition 6.7*** If the Banach space X is separable, then the set *

is a dense G_{δ }subset of .If X and Y are both separable, then the set

is a dense Gδ subset of .

Proof. Let W be a non-empty open subset of and U_{1}, U_{2} be nonempty open subsets of X. Pick . By the definition of ,there exists such that and is nilpotent. Let be such that .Clearly the space

has finite codimension in Y, while has finite codimension in X. Since b is a dual pairing of X and Y, we can find and such that .Consider the element

and let for .From the above properties of f_{j} and u_{j} it immediately follows that the range of is contained in the range Q of and that the restrictions of to Q coincide. Hence for any Thus all ^{T} _{ }are nilpotent and therefore all _{} belong to .Since W is open in we can pick close enough to zero to ensure that . Now from the definition of .it immediately follows that , .Thus

Summarizing the above, we have found such that .Using the fact that carries a topology stronger than the one defined by the injective norm, we see that the map from to L(X) satisfies (6.3). By Corollary 6.6, the set of for which is supercyclic and is hypercyclic is a dense G_{δ }subset of . The proof of the first part of the proposition is complete.

Applying the first part of the proposition to* (Y, X) *instead of* (X, Y)* with* b(x,y) *replaced by *b(y,x), *we obtain that if Y is separable, then the set of for which is supercyclic and is hypercyclic is a dense G_{δ}subset of . The second part now follows from the first part and the fact that the intersection of two dense G_{δ} subsets of a complete metric space is again a dense G_{δ} set.

**Proof of Theorems 1.19 and 1.21**

Let us consider the case where Y=X′, b(x,y)=y(x) and .In this case the map is an isometry and is exactly the operator norm closure of the set of finite rank nilpotent operators. Moreover, taking into account that for any , we see that Theorem 1.19 is an immediate corollary of Proposition 6.7 for this specific choice of M and b.

Recall that the projective norm on the tensor product X ⊗ Y of the Banach spaces X and Y is defined by the formula

where the infimum is taken over all possible representations of as a finite sum .If we consider the case and the completion of with respect to the projective norm, then is exactly the set of nuclear quasinilpotent operators, and we see that Theorem 1.21 is an immediate corollary of Proposition 6.7 for this choice of *M *and *b.*

**Proof of Theorem 1.6**

Let X ∈ M By Lemma 3.5, there exist sequences and in X and X′ respectively such that conditions (3.5.1–3.5.4) are satisfied. Since{f_{n}} is uniformly equicontinuous, we can pick a nonzero continuous seminorm p on X such that for any whenever . By Lemma 3.3, the closed balanced convex hull K of a Banach s-disk. That is, the Banach space X_{k} is separable. It is also clear from Lemma 3.3 that any has shape for some and ,where ||x|| is the norm of X in the Banach space X_{k}. Consider the bilinear form on X × *l _{1}* defined by the formula

It is easy to see that b is well-defined, continuous and for any x ∈ X and a∈ *l _{1}*. Moreover, b separates points of X

Using boundedness of {y_{n}} in X_{k} and {w_{n}} in l_{1}, summability of and the definition of b, we see that the right-hand side of the first equality in the above display defines a continuous linear operators T : X → X, taking values in X_{k}. Since the restriction of I+T to Xk is hypercyclic on Xk, Xk is dense in X and Xk carries the topology stronger than the one inherited from X, we see that I+T is hypercyclic (any hypercyclic vector for is also hypercyclic for I+T). Pick a hypercyclic vector a for . Since a ≠ 0 and b separates points of l_{1} , the functional b(⋅, a) is non-zero. Since a is hypercyclic for , we can pick a strictly increasing sequence of positive integers such that for any . Then

Let and . Clearly U and V are non-empty open subsets of since the functional b(⋅,a) is non-zero). Moreover, from the above display it follows that for each . Hence I+T is non-mixing. Since I+T is hypercyclic, the proof of Theorem 1.6 is complete.

**Proof of theorem 1.15**

Let X be an infinite dimensional locally convex space, such that both X and admit l_{1}-sequences with dense span. By Lemma 3.7, there exist sequences and in X and X' respectively, satisfying (3.7.1–3.7.3). By Lemma 3.3, the closed balanced convex hulls K and D of {x_{n}: n ∈ } and {f_{n} : n ∈} are Banach s-disks in X and respectively. Moreover, X_{K} is dense in X and X_{D}′ is dense in . Since D is β-compact, it is also σ (X′,X) -compact. Hence the seminorm on X is continuous with respect to the Mackey topology . Clearly each is bounded by 1 on and therefore {f_{n} : n ∈} is uniformly equicontinuous. By Lemma 3.7, we can assume that the same holds true if the original l_{1}-sequence in is uniformly equicontinuous. Assume for time being that either X carries the Mackey topology τ or the original l_{1}-sequence in is uniformly equicontinuous. Then {f_{n} : n ∈} is uniformly equicontinuous.

Consider the bilinear form on X x X′ defined by the formula β (x, f )= f (x) Clearly β separates points of X and X′ and β is separately continuous on . Since X_{K} is dense in X and is dense in , the bilinear form , being the restriction of β to , separates points of X_{K} and . Moreover separate continuity of β implies separate continuity of b and therefore continuity of b on by means of the uniform boundedness principle (every separately continuous bilinear form on product of Banach spaces is continuous). Let , for which the operator T_{ξ} defined in (6.8) is nilpotent. By Proposition 6.7, the set is a dense G_{δ} subset of . In particular, we can pick such that and are hypercyclic. Using once again the theorem characterizing the shape of elements of the projective tensor product, we see that there exist bounded sequences and in X_{K} and respectively and λ ∈ l_{1} such that . Then the operators T_{ξ}and S_{ξ} act according to the following formulae on X_{K} and respectively:

where we used the specific shape of our bilinear form. Using boundedness of {y_{n}} in X_{K} and {g_{n}} in and summability of , we see that the right-hand sides of the equalities in the above display define linear operators T and S on X and X′, taking values in X_{K} and respectively. Since is uniformly equicontinuous and {g_{n}} is bounded in , is also uniformly equicontinuous. It follows that T is continuous as an operator from X to X_{K} and therefore T ∈ L(X). It is also easy to verify that S = T′ and therefore Since the restriction I +T_{ξ} of I+T to X_{K} is hypercyclic on X_{K}, X_{K} is dense in X and X_{K} carries the topology stronger than the one inherited from X, we see that I+T is hypercyclic (any hypercyclic vector for I +Tξ is also hypercyclic for I+T). Similarly = is hypercyclic on (any hypercyclic vector for I + S_{ξ }is also hypercyclic for I+S). Hence I+T is a dual hypercyclic operator. In order to complete the proof of Theorem 1.15, it remains to consider the case when X carries the weak topology σ =σ (X,X′) . By the already proven part of Theorem 1.15, there is a dual hypercyclic operator R on X_{τ}. Since , R ∈ L(X). Since τ is stronger than σ, R is hypercyclic on X= X_{σ} . Hence R is dual hypercyclic on X. The proof of Theorem 1.15 is complete.

**Generic bilateral weighted shifts **

For each , T_{w }stands for the bounded linear operator acting on , 1≤ p < ∞ or , defined on the canonical basis by

If additionally w_{n}≠ 0 for each , the operator T_{w} is called the bilateral weighted shift with weight sequence w. Cyclic properties of **bilateral weighted shifts** have been intensely studied. Hypercyclicity and supercyclicity of bilateral weighted shifts were characterized by Salas [12,31] in terms of the weight sequences. It was observed in [37], Proposition 5.1 that the Salas conditions admit a simpler equivalent form:

**Theorem B** Let T be a bilateral weighted shift with weight sequence w acting on with 1≤ p < ∞ or . Then T is **hypercyclic** if and only if for any

and T is supercyclic if and only if for any (6.10)

(6.11)

where for with a ≤ b .

We address the issue of hypercyclicity and supercyclicity of generic bilateral weighted shifts in the Baire category sense. For each c > 0 let

Clearly B_{c} endowed with the coordinatewise convergence topology is a compact metrizable topological space.

**Theorem 6.8** Let 1≤ p < ∞ . For each c > 1 the set of w∈ B_{c} for which T_{w} acting on is hypercyclic is a dense G_{δ}- subset of B_{c}. For each c > 0 the set of w∈ B_{c} for which T_{w} is supercyclic and I+T_{w} is hypercyclic is a dense G_{δ}-subset of B_{c}.

It is worth noting that if w_{n}=0 for some n ∈ then the range of the operator T_{w} is not dense and therefore T_{w} can not be supercyclic. Thus any T_{w} for w from the dense G_{δ} -sets in the above theorem are indeed bilateral weighted shifts. Recall that T_{w} is compact if and only if w∈ . For compact bilateral weighted shifts we can replace the coordinatewise convergence topology on the space of weights by stronger topologies.

**Theorem 6.9 **Let 1≤ p < ∞ . Let E be a linear subspace of carrying its own F -space topology stronger than the one inherited from and such that the space ψ() of sequences with finite support is densely contained in E. Then the set of w∈E for which T_{w} acting on is supercyclic and I+T_{w} is hypercyclic is a dense G_{δ}-subset of E.

**Remark 6.10** As a corollary of the above theorem we obtain that the set of weights w∈ for which T_{w} acting on is supercyclic and I+T_{w} is hypercyclic is a dense G_{δ} -subset of the Banach space .

It is easy to see that the dual of T_{w} acting on for 1< p <∞ acts on defined by according to the formula

Considering the isometry U on defined by for each n ∈, we see that where for any n ∈. Thus is hypercyclic or supercyclic if and only if ′ (acting on is. In the case p =2, the Hilbert space adjoint T_{w}* acts on in a similar way and is unitarily similar to with a diagonal unitary operator providing the similarity. Thus the cyclicity properties of ′ and T_{w}* are the same.

Taking into account the fact that the map is a homeomorphism of B_{c} onto itself for each c > 0, we immediately obtain the following corollary of Theorem 6.8.

**Corollary 6.11** For each c>1 the set of w∈B_{c} for which both T_{w} and acting on are hypercyclic is a dense G_{δ}-subset of B_{c}. For each c > 0 the set of w∈ B_{c} for which both T_{w} and are supercyclic and both I + T_{w} and I +′ are hypercyclic is a dense G_{δ}-subset of B_{c}.

Similarly the next corollary follows from Theorem 6.9.

**Corollary 6.12** Suppose that the space E from Theorem 6.9 satisfies the additional symmetry condition that is an invertible continuous linear operator on E for some k∈ . Then set of w∈ E for which both T_{w} and acting on are supercyclic and both I+T_{w} and I + are hypercyclic is a dense G_{δ} -subset of E.

Applying the above corollary to weighted c_{0}-spaces with symmetric weight sequence yields the following result.

**Corollary 6.13** Let be any sequence of positive numbers. Then there exists w∈ such that |w_{n}|≤a_{|n|} for each n ∈ , T_{w} and acting on are supercyclic and I+T_{w} and I + are hypercyclic

We conclude this section by proving Theorems 6.8 and 6.9.

**Proof of Theorem 6.8**. It is straightforward to verify that the maps and from Ω= B_{c} into satisfy (6.3). Pick a non-empty open subset U of B_{c} and non-empty open subsets V and W of .

**Case c > 1:** By definition of the topology of B_{c}, there exist w∈U and a positive integer m such that w_{k}=c for k > m, w_{m}=c^{-1} for k <-m and w_{k} ≠ 0 for −m ≤ k ≤ m . According to Theorem B, the bilateral weighted shift T_{w }is hypercyclic. Hence we can choose x∈V and such that . Thus and therefore (6.4) is satisfied. By Theorem 6.2, the set of w∈ B_{c} for which T_{w} is hypercyclic is a dense G_{δ} -subset of B_{c}.

**Case c > 0:** As above, there exist w∈ U and a positive integer m such that w_{k}=c for k > m, w_{k}=c/2 for k <-m and w_{k} ≠ 0 for −m ≤ k ≤ m. According to Theorem B, the bilateral weighted shift T_{w} is supercyclic Hence we can choose x∈ V, and such that . Thus and therefore (6.6) is satisfied. By Theorem 6.3, the set of w∈ B_{c} for which T_{w} is supercyclic is a dense G_{δ}-subset of B_{c}.

Finally, we can pick m ∈ , w∈ U, x∈ V, and y∈ W, such that w_{m}= 0, w_{m} ≠ 0 for k > m, x and y have finite supports and x_{k} = y_{k} = 0 for k > m. It is straight forward to check that

By Corollary 6.6, the set of w∈ B_{c} for which I+T_{w} is hypercyclic is a dense G_{δ}-subset of B_{c}.

Proof of Theorem 6.9. We use the same notation as in the proof of Theorem 6.8. Since is dense in in E, we can choose w′∈U , x′∈V , y′∈W and a positive integer m such that x′_{k} = y_{k}′ =w_{k}′ = 0 if |k |>m . Choosing positive numbers ε_{n} for n >-m small enough, we can ensure that the series converges in E to w′′∈E , with w = w′ + w′′∈U and for k>-m. Clearly is a closed invariant subspace of Tw and x,y ∈F. Moreover, the restriction R of Tw to F is isometrically similar to the backward weighted shift on l_{p} with weight sequence By the Hilden and Wallen theorem [30] and the Salas theorem [12], R is supercyclic and I + R is hypercyclic. Since and we see that the open subsets V ∩F and W∩F of F are non-empty. Since R is supercyclic, we can choose x∈V ∩F , and such that . Thus and therefore (6.6) is satisfied. By Theorem 6.3, the set of w∈E for which T_{w} is supercyclic is a dense G_{δ}-subset of E. Finally, since I + R is hypercyclic, we can choose x∈V ∩F and such that . Thus and therefore (6.4) is satisfied for the map . By Theorem 6.2, the set of w∈E for which I+T_{w} is hypercyclic is a dense G_{δ}-subset of E.

**Mixing Operators on Spaces with Weak Topology**

In this section we shall prove Theorem 1.8. In order to do so we need a characterization of linear maps with no non-trivial finite dimensional invariant subspaces. The underlying field plays no role in this linear algebraic statement, so, for sake of generality we formulate and prove it for linear maps on linear spaces over an arbitrary field.

**Linear maps without finite dimensional invariant subspaces**

Throughout this subsection k is a field, X is a linear space over k, T: X→X is a k -linear map, P=k[z] is the space of polynomials on one variable, P=k(z) is the space of rational functions and M is the operator on P of multiplication by the argument:

We denote and consider the degree function extending the conventional degree of a polynomial. We set deg (0)= −∞ and let deg (p / q)= deg p − deg q , where P and q are non-zero polynomials and the degrees in the right hand side are the conventional degrees of polynomials. Clearly this function is well-defined and is a grading on P, that is, it satisfies the properties:

for any ;

for any ;

(d#) if and , then .

Note that if p is a non-zero polynomial, then deg P is the usual degree of P.

As usual, a linear subspace E of X is called T - invariant if and it is called T- biinvariant if and . The following lemma is a key ingredient in the proof of Theorem 1.8.

**Lemma 7.1** Let T be a linear operator on a linear space X with no non-trivial finite dimensional invariant subspaces and let L be a finite dimensional subspace of X. Then there exists such that p(T)(L)∩L = {0} for any with deg p ≥ n0 .

In order to prove the above lemma, we need some preparation.

**Lemma 7.2 **Let T be a linear operator on a linear space X. Then T has no non-trivial finite dimensional invariant subspaces if and only if p(T) is injective for any non-zero polynomial P.

**Proof.** If p is a non-zero polynomial and p(T) is non-injective, then there is non-zero x ∈ X such that p(T)x=0. Let k=deg p. It is straightforward to verify that is a nontrivial finite dimensional invariant subspace for T. Assume now that T has a non-trivial finite dimensional invariant subspac L. Let p be the characteristic polynomial of the restriction of T to L. By the Hamilton– Cayley theorem p(T) vanishes on L. Hence p(T) is non-injective.

**Definition 7.3** For a linear operator T on a linear space X we say that vectors x_{1},…, x_{n} in X are T - independent if for any polynomials p_{1},…, p_{n}, the equality implies p_{j}= 0 for 1≤ j ≤ n . Otherwise, we say that x_{1},…,x_{n} are T - dependent. A set A ⊂ X is called T - independent if any pairwise different vectors are T - independent.

For any non-zero x ∈ X, we define

**Lemma 7.4** Let T be a linear operator on a linear space X, x ∈ X \{0} and F(T, x)be the space defined in (29). Then F(T, x) is a T -biinvariant linear subspace of X.

**Proof. **Let y, u∈ F(T, x) and t,s∈k . Then we can pick and such that and . Hence . Since p1 and p2 are non-zero, the polynomial p1 p2 is also non-zero and we have and therefore F(T, x) is a linear subspace of X. Clearly , where . Hence Ty∈ F(T, x), which proves the T-invariance of F(T, x). Assume now that w∈X and Tw∈ F(T, x). Thus we can pick and such that . Hence , where , and therefore w∈ F(T, x). That is, F(T, x) is T-biinvariant.

By the above lemma, we can consider linear operators, being restrictions of T to the invariant subspaces F(T, x). The following lemma describes these restrictions in the case when T has no nontrivial finite dimensional invariant subspaces.

**Lemma 7.5 **Let T be a linear operator on a linear space X with no non-trivial finite dimensional invariant subspaces. Let also x ∈ X \{0} and F(T, x) be defined in (7.1). For each y∈ F(T, x) and any , satisfying , we write r_{x,y}= q/p. Then the rational function r_{x,y} does not depend on the choice of and satisfying , the map from F(T, x) to P is linear and for any y∈ F(T, x).

**Proof.** Let y∈ F(T, x) and , be such that and . Hence and therefore . Since x ≠ 0, Lemma 7.2 implies that , or equivalently, q/p=q_{1}/p_{1}. Thus q/p does not depend on the choice of and satisfying . In particular, the map from F(T, x) to P is well defined. Next, let y, u∈ F(T, x) and t,s∈k . Then we can pick and such that and . Hence . It follows that

and the linearity of the map is also verified. It remains to show that . Clearly . Since , we have . Hence .

**Lemma 7.6 **Let T be a linear operator with no non-trivial finite dimensional invariant subspaces acting on a linear space X and A be a T -independent subset of X. Then for each x∈A ,

where the spaces F(T, u) are defined in (7.1).

**Proof. **Assume that the intersection in the above display contains a non-zero vector u. Since u∈F(T, x) , there exist and such that . Since u ≠ 0 and p ≠ 0 , according to Lemma 7.2, p(T)u ≠ 0 . It follows that q ≠ 0 . On the other hand, since u is a non-zero element of the span of the union of F(T, Y) for y∈A\{x} , there exist pairwise different and such that . Pick and for which for 1≤ j ≤ n . Since u_{j} ≠ 0 and p_{j} ≠ 0 , from Lemma 7.2 it follows that and therefore . Consider the polynomials for 1≤ j ≤ n . Then

Taking into account that , we obtain

Since the polynomials and are non-zero, the last display contradicts the T -independence of A, since are pairwise different elements of A.

Proof of Lemma 7.1. Clearly there exists in L a maximal T -independent subset (since T -independence implies linear independence and L is finite dimensional, A is a finite set). It follows from the maximality of A that L is contained in the sum of F(T, x j ) for 1≤ j ≤ k . The last sum is direct according to Lemma 7.6:

Thus any x ∈ N can be uniquely presented as a sum , where . Using Lemmas 7.5 and 7.4, we can consider the linear operator

According to Lemmas 7.4 and 7.5 we also have that N is a T-biinvariant subspace of X and for 1≤ j ≤ k . For each x∈N , let

Clearly δ (0)= −∞ and δ (x) ∈ for each x∈N \{0} . Let also

Using the fact that L is finite dimensional, we will show that Δ^{+} and Δ^{-} are finite. Indeed, assume that either Δ^{+} = +∞ or Δ^{-} = −∞ . Then there exists a sequence of non-zero elements of L such that the sequence is strictly monotonic. For each l we can pick such that . Then there is such that the set is infinite. It follows that the degrees for are pairwise different. Property (d3) of the degree function implies that the rational functions for are linearly independent. Hence the infinite set is linearly independent in X, which is impossible since all u_{1} belong to the finite dimensional space L. Thus Δ^{+} and Δ^{−} are finite.

Now let and m= deg p . From (d1) and the equality , we immediately get that for each x∈N . Therefore, . In particular, if m> Δ^{+} − Δ^{−} , then

Thus δ (u) >δ (v) for any non-zero u∈P(T)(L) and v∈L , which implies that whenever . Thus the number satisfies the desired condition.

**Proof of Theorem 1.8**

The implications (1.8.4)⇒(1.8.3)⇒(1.8.2) are trivial. Assume that T is transitive and T′ has a non-trivial finite dimensional invariant subspace. Then T has a non-trivial closed invariant subspace of finite codimension. Passing to the quotient by this subspace, we obtain a transitive operator on a finite dimensional topological vector space. Since there is only one Hausdorff linear topology on a finite dimensional space, we arrive to a transitive operator on a finite dimensional Banach space. Since transitivity and hypercyclicity for operators on separable Banach spaces are equivalent [3], we obatin a hypercyclic operator on a finite dimensional Banach space. On the other hand, it is well known that such operators do not exist, see, for instance, [7]. This proves the implication (1.8.2)⇒(1.8.1) . It remains to show that (1.8.1) implies (1.8.4).

Assume that (1.8.1) is satisfied and (1.8.4) fails. Then there exist non-empty open subsets U and V of X and a sequence of polynomials such that and for each . Since X carries weak topology, there exist two finite linearly independent sets and in X′ and two vectors and such that and , where

Let . Since T′ has non non-trivial finite dimensional invariant subspaces, by Lemma 7.1, for any sufficiently large l , . For such an l, the equality together with the injectivity of , provided by Lemma 7.2, and the definition of L imply that the vectors are linearly independent. Hence there exists a vector u ∈ X such that

Since , the last display implies that and . Hence contains and therefore is non-empty. This contradiction completes the proof.

Spaces without Supercyclic Semigroups

We shall prove Theorem 1.7 and show that on certain topological vector spaces there are no strongly continuous supercyclic semigroups . In this section by the dimension dimX of a vector space X we mean its algebraic dimension (=the cardinality of the Hamel basis). Symbol c stands for the cardinality of continuum: . The next theorem is the main result of this section.

**Theorem 8.1 **Let X be an infinite dimensional locally convex space such that either in X or in there are no compact metrizable subsets whose linear span has dimension c. Then there are no strongly continuous supercyclic semigroups on X.

The above theorem immediately implies the following corollary.

**Corollary 8.2 **Let X be an infinite dimensional locally convex space such that dim X < c or . Then there are no strongly continuous supercyclic semigroups on X.

We prove Theorem 8.1 at the end of this section. First, we shall prove Theorem 1.7 by means of application of Theorems 1.8 and 8.1.

**Proof of Theorem1.7.** Let T ∈L(ω) be such that T′ has no nontrivial finite dimensional invariant subspaces and be a sequence of polynomials such that as . Since the topology of ω is weak, Theorem 1.8 implies that for each non-empty open subsets U and V of ω, for all sufficiently large l. Hence is dense in ω×ω. By Theorem is universal. It remains to show that there are no strongly continuous supercyclic semigroups on X. Recall that and . Thus Corollary 8.2 implies that there are no supercyclic strongly continuous operator semigroups on ω.

The rest of the section is devoted to the proof of Theorem 8.1. We need some preparation.

**Lemma 8.3 **Let X be a finite dimensional topological vector space of the real dimension >2. Then there is no supercyclic strongly continuous operator semigroup on X.

**Proof.** As well-known, any strongly continuous operator semigroup on has shape , where such that the semigroup is supercyclic and n ≥ 3 if , n ≥ 2 if . Since the operators are invertible and commute with each other, Proposition G implies that the set W of universal elements for the family is dense in . On the other hand, for each c > 0 and any x ∈ , from the restrictions on n it follows that the closed set is nowhere dense in (one can use smoothness of the map to see that the topological dimension of is less than that of . Hence, each x∈W is universal for the family for any c > 0 . Now if (a,b) is a finite subinterval of (0,∞) , it is easy to see that the family contains for a sufficiently large c > 0. It follows that for each x ∈ W the set is dense in . Taking into account that (a,b) is arbitrary and W is dense in , we see that is dense in . Applying Theorem 6.3, we see that for a generic t ∈ in the Baire category sense the operator e^{tA} is supercyclic. This contradicts the well-known fact that there are no supercyclic operators on finite dimensional spaces of real dimension >2.

The following lemma appears as Lemma 2 in reference [38]. It is worth noting that under the Continuum Hypothesis its statement becomes trivial.

**Lemma 8.4 **Let (M, d) be a separable complete metric space and X be a topological vector space. Then for any continuous map f :M →X , the algebraic dimension of spanf(M) is either finite or countable or continuum.

We use the above lemma to prove the following dichotomy.

**Lemma 8.5** Let be a strongly continuous operator semigroup on a topological vector space X and x ∈ X. Then the space is either finite dimensional or has dimension c.

**Proof.** From Lemma 8.4 it follows that dimC(x) is either finite or or c. Thus it suffices to rule out the case

Assume that . Restricting the operators T_{t} to the invariant subspace C(x) , we can without loss of generality assume that C(x) = X . Thus and therefore X is the union of an increasing sequence of finite dimensional subspaces. For each ε > 0 let . First, we shall show that each X_{ε} is finite dimensional. Let ε > 0, 0 <α <ε and for . Clearly A_{n} are closed subsets of the interval [α ,ε ] and since X is the union of E_{n}. By the Baire category theorem there is such that A_{n} has non-empty interior in [α ,ε ] . Hence we can pick a,b ∈ such that α ≤ a < b ≤ε and T_{t}x∈E_{n} for any t∈[a,b] . We shall show that . Assume, it is not the case. Then the number belongs to [b,∞) . Since the set is closed, . Since [a,b] is uncountable and the span of is finite dimensional, we can pick and such that

Since by definition of c, we can pick t such that . Since t > c ≥ t_{n} , formula (8.1) implies that

because for . This contradiction proves that . . Thus X_{ε} is finite dimensional for each ε > 0. Since , it follows that T_{t} has finite rank for any t > 0.

Now assume that t > 0. Since Tt has finite rank, is a closed subspace of X of finite codimension. It is also clear that F_{t} is T_{s}-invariant for each s ∈ Passing to quotient operators, S_{s} ∈L(X /F_{t} ), , we arrive to a strongly continuous semigroup on the finite dimensional space X / F_{t} . Hence there is such that for any s ∈ . Thus each S_{s} is invertible. Since each Ss is a quotient of Ts, we see that

It follows that T_{t} for t > 0 have the same rank k ∈ . Passing to the limit as t → 0, we see that the identity operator I=T_{0} is the strong operator topology limit of a sequence of rank k operators. Hence rk I ≤ k . That is, X is finite dimensional. This contradiction completes the proof.

**Lemma 8.6 : **

*Let X be a topological vector space in which the linear span of each metrizable compact subset has dimension <c. Then for any strongly continuous operator semigroup* *on X and any *x ∈ X, *the space* *is finite dimensional*.

**Proof:** Let be a strongly continuous operator semigroup on X and x ∈ X. Strong continuity of implies that for any , the set is compact and metrizable. Hence for any , where . Since the sum of countably many cardinals strictly less than c is strictly less than c, we see that

By Lemma 8.5, is finite dimensional.

**Proposition 8.7: **Let X be an infinite dimensional locally convex space such that in the span of any compact metrizable subset has dimension <c. Then there is no strongly continuous supercyclic operator semigroup on X.

**Proof:** Assume that there exists a supercyclic strongly continuous operator semigroup on X. It is straightforward to verify that is a strongly continuous semigroup on . Pick three linearly independent vectors f1, f2 and f3 in X′. By Lemma 8.6, is finite dimensional for . Clearly each *E _{j}* -invariant for any Then is finite dimensional and -invariant for any Since

**Proof of Theorem 8.1:** If X has no compact metrizable subsets whose linear span has dimension c, Lemma 8.6 implies that the linear span of any orbit is finite dimensional. It follows that is not supercyclic. It remains to consider the case when has no compact metrizable subsets whose linear span has dimension c and apply Proposition 8.7.

**The space **φ

Recall that φ is a linear space of countable algebraic dimension carrying the strongest locally convex topology. In this section we mention certain properties of φ, mainly those which are related to continuous linear operators. It is well known [17] that φ is complete and all linear subspaces of φ are closed. Moreover, infinite dimensional subspaces of φ are isomorphic to φ. It is also well-known that for any topology θ on φ such that (φ,θ) is a topological vector space, θ is weaker than the original topology of φ. The latter observation immediately implies the following lemma.

**Lemma 9.1: ***For any topological vector space X and any linear map T is continuous.*

**Lemma 9.2:** Let X be a topological vector space and be a surjective continuous linear operator. Then X is isomorphic to

**Proof.** Since T is linear and surjective, there exists a linear map * *such that TS=I. By Lemma 9.1, S is continuous. Consider the linear maps

It is easy to see that A and B are continuous and that AB=I and BA=I. Hence B is a required isomorphism.

**Corollary 9.3:** *Let X be a topological vector space. Then the following conditions are equivalent. pt*

(9.3.1) X is isomorphic to a space of the shape Y × φ,, where Y is a topological vector space ;

(9.3.2) X has a quotient isomorphic to φ;

(9.3.3) there is * *such that T(X) is infinite dimensional.

**Proof:** The implications (9.3.1)⇒(9.3.2)⇒(9.3.3) are trivial. Assume that (9.3.3) is satisfied. That is, there is with infinite dimensional T(X). Since any infinite dimensional linear subspace of φ is isomorphic to φ, we see that T(X) is isomorphic to φ. Hence there is a surjective * *By Lemma 9.2 X is isomorphic to Y × φ, where Y = ker S . Hence (9.3.3) implies (9.3.1), which completes the proof.

**Cyclic operators on φ**

Clearly φ is isomorphic to the space *p* of all polynomials over *k* endowed with the strongest locally convex topology. The shift operator on φ is obviously similar to the operator

(9.1)

For each we denote

* *(9.2)

Clearly *p*_{n} is an n-dimensional subspace of *p*.

**Lemma 9.4: ***An operator* * is cyclic if and only if T is similar to the operator M.*

**Proof:** Clearly 1 is a cyclic vector for M. Hence any operator similar to M is cyclic. Now let be cyclic and * *be a cyclic vector for T. Then the vectors* *are linearly independent. Indeed, otherwise their span is finite dimensional, which contradicts cyclicity * *is an algebraic basis of φ. It is easy to see then that the linear map is invertible and . By Lemma 9.1, J and J^{-1} are continuous. Hence T is similar to M.

We need a multicyclic version of the above lemma.

**Lemma 9.5: **Let * . Then the following conditions are equivalent.*

(9.5.1) T is multicyclic;

(9.5.1) there exists k∈ and a linear subspace Y of φ of finite codimension such that * *and the restriction * i*s similar to *M ^{k}*, where M is the operator defined in (9.1).

**Proof:** First, assume that (9.5.2) is satisfied. Pick a finite dimensional subspace Z of such that *. *Since * *is similar to *M ^{k}*,, we can pick an invertible linear operator such that

Hence the linear span of the union of * *for * *contains *.*Thus T is m -cyclic with m=dimL. The implication (9.5.2)⇒(9.5.1) has been verified.

Assume now that T is n -cyclic for some . Then there is an n -dimensional subspace L of φ such that

* (9.3)*

(again, we use the fact that any linear subspace of φ is closed and therefore a dense subspace of φ must coincide with φ). We will use the concept of T -independence, introduced in Section 7. Since T -independence implies linear independence, any T -independent subset of L has at most n elements. Let k be the maximum of cardinalities of T -independent subsets of L and A be a T -independent subset of cardinality k. Since A is linearly independent, we can pick a subset * *of cardinality n-k such that * *is a basis in L. From the definition of k it follows that for any is not T independent and therefore using T -independence of A, we can find polynomials *p _{b}* and

* *(9.4)

Now let m=0 if * *and otherwise. Consider the spaces

Then Z is finite dimensional and . Obviously,

Let * *and *. *Since * *and * , *we can find polynomials q,r such that and * . *Then

*.* Since* .* According to (9.4), *. *Since Y is invariant for *. *Thus

(9.6)

Since * *is a basis of L, from (9.5) and (9.6) it follows that for each .According to (9.3), . Since Z is finite dimensional, we see that Y has finite codimension in φ. In particular, Y is non-trivial and therefore . That is, 1 ≤ k ≤ n and . Now consider the linear operator *, *which sends the monomial * , *where* *and * *are uniquely defined by the equation. By definition of Y, J is onto. From T -independence of A it follows that J is also one-to-one. By definition of J, we have *. . *Hence * *for any *. *That is, *M ^{k}* and are similar.

**Corollary 9.6:** *Let T be a multicyclic operator on φ. Then T is not onto.*

**Proof : **According to Lemma 9.5, we can decompose φ into a direct sum * , *where Z has finite dimension * , *and * *is similar to *M ^{k}* for some

**Proof of Theorem 1.12**

In this section *X* is a topological vector space, which has no quotient isomorphic to φ. We have to show that there are no cyclic operators with dense range on *X×φ*. Assume the contrary and let be a cyclic operator with dense range. Consider the matrix representation of T:

With T acting according to the formula * *Since T is cyclic, we can pick a vector such that is dense in* X×φ. *Since T has dense range, then Tm has dense range for any . Thus is dense in *X×φ* for each . Let for and* uk =φ. *Since is dense in X×φ, we see that is dense in φ for any . Hence F_{m}= φ for any . Since X has no quotients isomorphic to φ, Lemma 9.3 implies that is a finite dimensional subspace of φ. Then the space

is also finite dimensional. Clearly and * *for any *.*It follows that each uk belongs to the space spanned by the union of for . Since L is finite dimensional, D is multicyclic. By Lemma 9.5, we can decompose φ into a direct sum *, *where Z is finite dimensional, * *and * *is similar to *M _{n}* for some . That is, there exists an invertible

**Case 1:** The sequence is bounded from above. In this case is finite dimensional. Since J is invertible, is finite dimensional. Since P has finite dimensional kernel, * *is finite dimensional. We have arrived to a contradiction with the equality* F _{0}=φ.*

**Case 2: **The sequence * *is unbounded from above. Since * *is finite dimensional,

We shall show that * *whenever * .* Indeed, let be such that *.* By definition of P, and . As we know, . Hence . Since belong to, Y we have . Thus there is such that . Hence

Since *, *we have (the degree of the sum of two polynomials of different degrees equals to the maximum of the degrees). Since the sequence is unbounded from above, there is such that and according to the just proven statement, we will have * *for j ≥ k. Since any family of polynomials with pairwise different degrees is linearly independent, we see that the vectors *JPu _{j}* for j ≥ k are linearly independent. Since JP is a linear operator, the vectors

We shall prove the following lemma, which is a key for the proof of Theorem 1.13.

**Lemma 10.1:** *Let be a sequence of infinite dimensional locally convex spaces such that*

*(10.1.1) there exists a dense linear subspace Y of X _{0}, carrying a topology, stronger than the one inherited from X_{0} and turning Y into a separable metrizable topological vector space;*

*(10.1.2) there exists with dense range;*

*(10.1.3) for each , there exists with dense range.*

*Then there is a hypercyclic operator S on *

It is worth noting that condition (10.1.1) implies that* X _{0}* is separable, condition (10.1.2) implies that

**Lemma 10.2: **Let X and Y be topological vector spaces such that there exists *with dense range. Then for any closed hyperplane H of X, there exists S ∈ L(H,Y) with dense range.*

**Proof:** Let * *be the restriction of *T* to H. We can write *, *where and *. *If *T _{0}* has dense range, thenis a continuous linear operator from H to Y with dense range. It remains to consider the case when the range of

**Proof of Lemma 10.1:** Let * *be a base of topology of Y and

Clearly X is the union of the increasing sequence of subspaces *Z _{n}* and each

(a1)

(a2)

(a3)

(a4)

(a5)

By (10.1.2) there exists with dense range. Since Y is dense in *X _{0}=Z_{0}, S_{0}* has dense range and

The operator *S _{n}* is continuous since

Condition (a2) ensures that there is a unique operator S∈ L(X) such that * *for any. From (a4) it now follows that

According to the above display, the set is contained in the orbit *.* By (a5) A is dense in *Y*. Since *Y* is dense in* X _{0}* and carries a stronger topology, A is dense in

**Remark 10.3: ***The orbit of the hypercyclic vector constructed in the proof of Lemma 10.1 is not just dense. It is sequentially dense. The latter property is strictly stronger than density already for countable direct sums of separable infinite dimensional Banach spaces.*

**Proof of Theorem 1.14**

Let* *for each * *and *.*We shall apply Lemma 10.1. Condition (10.1.1) is satisfied according Lemma 3.3. Indeed *X _{1 }*admits a Banach s -disk with dense span. By Lemma 3.8, spaces

**Proof of Theorem 1.13**

As we have already mentioned, s Fréchet space *X* belongs to if and only if X is infinite dimensional, separable and non-isomorphic to ω. Moreover, in any separable Fréchet space there is an -sequence with dense span.

**Lemma 10.4:** Let X and Y be separable infinite dimensional Fréchet spaces. Then the following conditions are equivalent

(10.4.1) there is no *T ∈ L(X, Y) *with dense range;

(10.4.2) X is isomorphic to ω and Y is not isomorphic to Ω.

**Proof:** If both X and Y are isomorphic to ω, then obviously there is a surjective *T ∈ L(X,Y)*. If X is isomorphic to ω, *Y* is not and *T ∈ L(X,Y)*, then *Z=T(X) *carries minimal locally convex topology [29] since ω does. It follows that *Z* is either finite dimensional or isomorphic to ω and therefore complete. Hence *Z* is closed in *Y*. It follows that since Y is neither finite dimensional nor isomorphic to . Thus there is not *T ∈ L(X,Y)* with dense range. It remains to show that there is *T∈ L(X,Y)* with dense range if X is not isomorphic to ω. In this case the topology of X is not weak and it remains to apply Lemma 3.6.

**Lemma 10.5:** *Let X be the countable locally convex direct sum of a sequence of separable Fréchet spaces infinitely many of which are infinite dimensional. Then one of the following two possibilities occurs:*

(10.5.1) *X is isomorphic to , where Y is a separable infinite dimensional Fréchet space and Z is the locally convex direct sum of an infinite countable number of copies of ω;*

(10.5.2) *X is isomorphic to , where each Y _{n} is a separable infinite dimensional Fréchet space non-isomorphic to ω.*

**Proof:** Separating the finite dimensional spaces, spaces isomorphic to ω and infinite dimensional spaces non-isomorphic to ω, we see that

where the sets A, B and C are pairwise disjoint, Xα is isomorphic to ω for each * *is a separable infinite dimensional Fréchet space nonisomorphic to ω for any *, X _{γ} *is finite dimensional for each

If B and C are finite, then A is infinite. Pick *. *Then

Clearly *Y* is a separable infinite dimensional Fréchet space. Since each *X _{α}* is isomorphic to ω, we fall into the case (10.5.1). If

Again Y is a separable infinite dimensional Fréchet space. Since each * *is isomorphic to ω, (10.5.1) is satisfied. If B is infinite and is infinite, then we enumerate both B and by the elements of +: .We arrive to

Since each *Xρ* is a separable infinite dimensional Fréchet space non-isomorphic to *ω*, (10.5.2) is satisfied. Finally, if *B* is infinite and is finite, we fix * *and write

Again *Z* and each *X _{β}* are separable infinite dimensional Fréchet spaces non-isomorphic to ω and we fall into the case (10.5.2).

We are ready to prove Theorem 1.13. Let X be a countable infinite direct sum of separable Fréchet spaces. If all the spaces in the sum, except for finitely many, are finite dimensional, then X is isomorphic to *Y×φ*, where Y is a Fréchet space. According to Theorem 1.12, X admits no cyclic operator with dense range. In particular, there is no supercyclic operator on X. If there are infinitely many infinite dimensional spaces in the sum defining X, then according to Lemma 10.5, we see that X is isomorphic to

where *Yn* are all separable infinite dimensional Fréchet spaces and either all *Yn* are non-isomorphic to ω or all *Yn* for are isomorphic to . In any case from Lemma 10.4 it follows that there exists with dense range and for each *,* there exists with dense range. By Lemma 10.1, there is a hypercyclic operator on X. The proof of Theorem 1.13 is complete.

The following lemma is a main tool in the proof of Theorem 1.11.

**Lemma 11.1: ***Let a locally convex space be the union of an increasing sequence of its closed linear subspaces. Assume also that for any there is an -sequence with dense span in X _{n} and the topology of is not weak, where . Then there exists a linear map such that for any , and the orbit is dense in X.*

Note that we do not claim continuity of the above operator S on *X*. Although if, for instance, *X* is the inductive limit of the sequence {*X*_{n}}, then continuity of S will immediately follow from the continuity of the restrictions *.*

**Proof of Lemma 11.1:** For each , let be an -sequence with dense span in Xn. For any , we apply Lemma 3.1 with the triple of spaces being (*X,X _{n},X_{n−1}*) to obtain a sequence in such that is uniformly equicontinuous, each

(p1)

(p2)

(p3)

(p4)

(p5)

Consider the linear map * *defined by the formula

Since is an -sequence in *X _{2} *and is uniformly equicontinuous, the above display defines a continuous linear operator from

The above series converge since * *is an -sequence and is uniformly equicontinuous. Applying the above property to the pair , we find such that

Consider now the linear map * *defined by the formula

The above display defines a continuous linear operator since

is an -sequence and is uniformly equicontinuous. From the last three displays it follows that and . From definition of w and u and the relation it follows that (p3) and (p4) for k=n are satisfied. Clearly (p5) for k=n is also satisfied. Since each vanishes on *X _{n-1}*, we have from the last display that for any

Condition (p2) ensures that there is a unique linear map such that * *for any *.* From (p4) it now follows that for each . Thus the set is contained in the orbit . By (p5) A is dense in Xk and therefore is dense in *X _{1}.* By (p2) is dense in

Before proving Theorem 1.11, we need to make the following two elementary observations.

**Lemma 11.2:*** Let X be an LB-space and Y be a closed linear subspace of X. Then either X /Y is finite dimensional or the topology of X /Y is not weak.*

**Proof:** Since X is an LB-space, it is the inductive limit of a sequence * *of Banach spaces. If *X /Y* is infinite dimensional, we can find a linearly independent sequence in X′ such that each fn vanishes on Y. Next, we find a sequence of positive numbers converging to zero fast enough to ensure that as for each *.*It follows that the sequence * *is pointwise convergent to zero on X. Since any LB-space is barrelled [17,29], the set is uniformly equicontinuous. Hence is a continuous seminorm on X. Since each *f _{n}* vanishes on

**Lemma 11.3:** *Let X be an inductive limit of a sequence of Banach spaces such that X _{0} is dense in X. Then X has no quotients isomorphic to *φ

**Proof: **Assume that X has a quotient isomorphic to *φ*. By Lemma 9.2 then X is isomorphic to *Y×φ* for some closed linear subspace Y of X. Let be the natural embedding. Since *X _{0}*is dense in X, J has dense range. Hence ′ is injective. Since X is isomorphic to

**Proof of Theorem 1.11**

Throughout this section X is the inductive limit of a sequence of separable Banach spaces. Let also be the closure of X_{n} in X. First, we shall prove the implication. Assume that (1.11.4) is satisfied. Then we can pick a strictly increasing sequence of non-negative integers such that for each , Hence, for any , we can pick a non-trivial finite dimensional subspace Y_{k} of such that . Thus the vector space X can be written as an algebraic direct sum

(11.1)

Apart from the original topology τ on X, we can consider the topology θ, turning the sum (11.1) into a locally convex direct sum. Obviously . On the other hand, if W is a balanced convex θ-neighborhood of 0 in X, then is a τ-neighborhood of zero in for any . Indeed, it follows from the fact that where and Z_{k} is finite dimensional. Since the topology of each is stronger than the one inherited from X, we see that is a neighborhood of zero in for each . Since X is the inductive limit of the sequence W is a τ -neighborhood of zero in X. Hence . Thus τ and therefore X is isomorphic to where Y is the locally convex direct sum of Yk for . Since Y_{k} are finite dimensional, Y is isomorphic to φ. Since is the inductive limit of the sequence of separable Banach spaces (with the topology inherited from ), the first one of which is dense, we see that (1.11.3) is satisfied. The implication is verified.

Assume now that (1.11.3) is satisfied. By Lemma 11.3, Y has no quotients isomorphic to φ. Theorem 1.12 implies now that there are no cyclic operators with dense range on X, which proves the implication . The implication is obvious since any hypercyclic operator is cyclic and has dense range. It remains to show that (1.11.1) implies (1.11.4). Assume the contrary. That is, (1.11.1) is satisfied and (1.11.4) fails. The latter implies that either there is n ∈_{+ }such that is dense in X or there is a strictly increasing sequence of non-negative integers such that is infinite dimensional for each . In the first case, it is easy to see that X ∈ M and therefore there is a hypercyclic operator on X by Corollary 1.4. We have obtained a contradiction with (1.11.1). It remains to consider the case when there exists a strictly increasing sequence of non-negative integers such that is infinite dimensional for each . By Lemma 11.2, the topology of each is not weak. Let . Since is a separable Banach space, there is an _{1}-sequence in with dense span. Since the topology on inherited from X is weaker than the Banach space topology of , is an _{1}-sequence with dense span in . By Lemma 11.1, there exists a linear map and such that for any the restriction of S to is continuous and the orbit is dense in X. Since the topology of is stronger than the one inherited from X, we have that each restriction of S to is a continuous linear operator from to X. Since X is the inductive limit of the sequence , is continuous. Hence S is a hypercyclic continuous linear operator on X. The existence of such an operator contradicts (1.11.1). The proof of the implication and that of Theorem 1.11 is now complete.

**Remarks on mixing versus hereditarily hypercyclic**

We start with the following remark. As we have already mentioned, φ supports no supercyclic operator [11], which follows also from Theorem 1.12. On the other hand, φ supports a transitive operator [22]. The latter statement can be easily strengthened with the help of Corollary 5.1. Namely, take the backward shift T on φ. That is Te_{0}=0 and for n ≥ 1, where is the standard basis in φ. Clearly T is a generalized backward shift and therefore T is an extended backward shift. By Corollary 5.1, I+T is mixing. Thus we have the following proposition.

**Proposition 12.1** φ* supports a mixing operator and supports no supercyclic operators.*

On the other hand, a topological vector space of countable algebraic dimension can support a hypercyclic operator, as observed by several authors, [22], for instance. The following proposition formalizes and extends this observation.

**Proposition 12.2*** Let X be a normed space of countable algebraic dimension. Then there exists a hypercyclic mixing operator T∈ L(X).*

**Proof. **Let be the completion of X. Then is a separable infinite dimensional Banach space. By Corollary 1.4, there is a hereditarily hypercyclic operator on . be a hypercyclic vector for S and E be the linear span of the orbit of x: . Grivaux [39] demonstrated that for any two countably dimensional dense linear subspaces E and F of a separable infinite dimensional Banach space Y, there is an isomorphism J :Y →Y such that J(E)= F . Hence there is an isomorphism such that J(X)= E . Let now . Since J(X)= E and E is S -invariant, X is T_{0}-invariant. Thus the restriction T of T_{0} to X is a continuous linear operator on X. Moreover, since the S -orbit of s is dense in the T_{0}-orbit of J^{-1}x is dense in . Since , the latter orbit is exactly the T -orbit of J^{-1}x and therefore J^{-1}x is hypercyclic for T. Hence T is hypercyclic. Next, T_{0} is mixing since it is similar to the mixing operator S. Hence T is mixing as a restriction of a mixing operator to a dense subspace.

By Proposition 1.1, if X is a Baire separable and metrizable topological vector space, then any mixing T∈ L(X) is hereditarily hypercyclic. From the above proposition it follows that there are mixing operators on countably dimensional normed spaces. The next theorem however implies that there are no hereditarily hypercyclic operators on countably dimensional topological vector spaces, emphasizing the necessity of the Baire condition in Proposition 1.1.

**Theorem 12.3** Let X be a topological vector space such that there exists a hereditarily universal family . Then . Proof. Since the topology of any topological vector space can be defined by a family of quasinorms [17], we can pick a nonzero continuous quasinorm P on X. That is, is nonzero, continuous, for any x,y ∈X, if x ∈ X , z ∈ K, | z |=1 and (X,τ _{p}) is a (not necessarily Hausdorff) topological vector space, where τp is the topology defined by the pseudometric d(x, y) = p(x − y) . The latter property implies that for any t ∈K and any sequence in X such that Let κ be the first uncountable ordinal (commonly denoted ω_{1}). We shall construct inductively sequences and of vectors in X and subsets of Z_{+} respectively such that for anyα < κ, pt

(s1) Aα is infinite and xα is a universal vector for the family ;

(s2) as n→∞ , n ∈ A_{α} for any β < α;

(s3) A_{α} \ A_{β} is finite for any β < α.

For the basis of induction we take A_{0}=_{+} and x_{0} being a universal vector for the family {T_{n} : n ∈_{+}}. It remains to describe the induction step. Assume that γ < κ and x_{α}, A_{α} satisfying (s1–s3) for α < γ are already constructed. We have to construct xγ and Aγ satisfying (s1–s3) for α=γ.

**Case 1: **γ has the immediate predecessor. That is γ =ρ +1 for some ordinal ρ < κ. Since x_{p} is universal for : , we can pick an infinite subset A_{γ} ⊂ A_{ρ} such that as n→∞ , n∈A_{γ} . Since A_{γ}is contained in A_{p}, from (s3) for α ≤ ρ it follows that A_{γ} \A_{β} is finite for any β < γ. Hence (s3) for α=γ is satisfied. Now from (s3) for α=γ. and (s2) for α < γ. it follows that (s2) is satisfied for α=γ. Next, since {T_{n} : n ∈_{+}} is hereditarily universal, we can pick x_{γ} ∈ X universal for {T_{n} : n ∈ A_{γ}}. Hence (s1) for α=γ. is also satisfied.

**Case 2:** γ is a limit ordinal. Since γ is a countable ordinal, we can pick a strictly increasing sequence of ordinals such that Now pick consecutively n_{0} from , n_{1} > n_{0} from , n_{2} > n_{1} from etc. The choice is possible since by (s3) for α < γ, each is infinite. Now let . Since , A_{α} is finite for each j ∈_{+}. Now if β < γ, we can pick j ∈_{+} such that β < α_{j} < γ. Then is finite by (s3) with α=j. Moreover, since A_{γ} is contained in up to a finite set, from (s2) with α=α_{j} it follows that p(T x )→0 as n→∞ , n∈A_{γ} . Hence (s2) and (s3) for α=γ are satisfied. Finally, since {T_{n} : n ∈_{+}} is hereditarily universal, we can pick xγ ∈ X universal for {Tn : n ∈A_{γ}}. Hence (s1) for α=γ is also satisfied. This concludes the construction of {x_{a}}_{α<κ} and {A_{a}}_{α<κ }satisfying (s1–s3).

In order to prove that , it suffices to show that vectors {x_{a}}_{α<κ} are linearly independent. Assume the contrary. Then there are n ∈N, and ordinals α_{1}<…<α_{n}<κ such that . By (s2) with α=α_{n}, we see that as k →∞ , for 1 ≤ j < n. Denoting for 1 ≤ j < n and using linearity of T_{k}, we obtain for any k ∈_{+}. Since p is a quasinorm, we have

The above display contradicts universality of , which is (s1) with α= α_{n}. This contradiction completes the proof.

**Corollary 12.4 ***A topological vector space of countable algebraic dimension supports no hereditarily hypercyclic operators.*

It is worth noting that there are infinite dimensional separable normed spaces, which support no supercyclic or transitive operators. We call a continuous linear operator T on a topological vector space X *simple* if T has shape T = zI + S , where z ∈ and S has finite rank. Observe that a simple operator on an infinite dimensional topological vector space is never transitive or supercyclic. Indeed, let T be a simple operator on an infinite dimensional topological vector space and λ ∈K, S∈ L(X) be such that T =λ I + S and S has finite rank. Then L=S(X) is finite dimensional. Since X/Y is infinite dimensional, we can pick non-empty open subsets U_{0} and V_{0} of X/L such that does not intersect . Let and . Clearly U and V are non-empty open subsets of X. Using the equalities T =λ I + S and S(X)=L, it is easy to see that for any . Hence T is non-transitive. Moreover since U and V are stable under multiplication by non-zero scalars, the projective orbit of any x ∈ U does not meet V. Hence U contains no supercylic vectors for T. Since the set of supercyclic vectors of any continuous linear operator is either dense or empty, T is non-supercyclic.

We say that a topological vector space X is* simple *if it is infinite dimensional and any T∈ L(X) is simple. Thus simple topological vector spaces support no supercyclic or transitive operators. Various examples of simple separable infinite dimensional normed spaces can be found in the literature [38,40-44]. Moreover, according to Valdivia [40], in any separable infinite dimensional Fréche spaced there is a dense simple hyperplane. All the examples of this type existing in the literature with one exception [44] are constructed with the help of the axiom of choice and the spaces produced are not Borel measurable in their completions. In [44] there is a constructive example of a simple separable infinite dimensional pre-Hilbert space H which is a countable union of compact sets.

Finally recall that an infinite dimensional topological vector space X is called rigid if L(X) consists only of the operators of the form I for λ ∈. Of course, a rigid space can not be locally convex. Clearly there are no transitive or cyclic continuous linear operators on a rigid topological vector space. Since there exist rigid separable spaces [45], we see that there are separable infinite dimensional spaces on which there are no cyclic or transitive operators.

**Concluding remarks and open problems**

We start by observing that the following questions remain open.

**Problem 13.1** *Is there a hereditarily hypercyclic operator on a countable direct sum of separable infinite dimensional Banach spaces?*

**Problem 13.2 ***Is there a hypercyclic strongly continuous operator semigroup on a countable direct sum of separable infinite dimensional Banach spaces?*

The most of the above results rely upon the underlying space being locally convex or at least having plenty of continuous linear functionals and for a good reason. As mentioned in the previous section, there are separable infinite dimensional spaces on which there are no cyclic or transitive operators. On the other hand, the absence of non-zero continuous linear functionals on a topological vector space does not guarantee the absence of hypercyclic operators on it. It is well-known [45] that the spaces L_{p}[0,1] for 0 ≤ p <1 are separable spaces having no non-zero continuous linear functionals. Ansari [9] raises a question whether these spaces support hypercyclic operators. Theorem 1.18 provides an easy answer to this question. Namely, consider the operator T ∈L(L_{p}[0,1]) , Tf (x)= f (x / 2) . It is straightforward to see that T is onto and has dense generalized kernel. Thus I+T is hereditarily hypercyclic according to Corollary 5.2.

It is obvious that an extended backward shift has dense range and dense generalized kernel. Unfortunately, the converse is not true in general. This leads naturally to the following question.

**Problem 13.3 ***Let T be a continuous linear operator on a separable Banach space, which has dense range and dense generalized kernel. Is it true that I+T is mixing or at least hypercyclic?*

From Corollary 5.1 and Corollary 2.14 it follows that if T is an extended backward shift on a separable infinite dimensional Banach space X, then both I+T and e^{T} are hereditarily hypercyclic. This reminds of the following question raised by Bermúdez, Bonilla, Conejero and Peris in reference [15].

**Question B2CP. ***Let X be a complex Banach space and T∈ L(X) be such that its spectrum σ (T) is connected and contains 0. Does hypercyclicity of I+T imply hypercyclicity of eT? Does hypercyclicity of eT imply hypercyclicity of I+T?*

We show that the answer to both parts of the above question is negative. Before doing this we would like to raise a similar question, which remains open. I+T

**Problem 13.4** *Let X be a Banach space and T∈ L(X) be quasinilpotent. Is hypercyclicity of I+T equivalent to hypercyclicity of e ^{T}?*

First, we introduce some notation. Let be the Hardy Hilbert space on the unit disk and be the space of bounded holomorphic functions . For any , the multiplication operator is a bounded linear operator on . It is also clear that . if is the adjoint of M_{φ}, then is the reflection of σ (M_{φ} ) with respect to the real axis. The following proposition is a direct consequence of a theorem by Godefroy and Shapiro [27][Theorem 4.9].

**Proposition 13.5*** Let . Then is hypercyclic if and only if*

(13.1)

The above Proposition calls for the following comment. A bounded linear operator T on a separable infinite dimensional Banach space X is said to satisfy the Kitai Criterion [46,47] if there exist dense subsets E and F of X and a map S : F →F such that TSy = y for any y∈F , T^{n}x→0 and as n→∞ for any x∈E and y∈F . As it is shown in [46,47], any operator satisfying the Kitai Criterion is hypercyclic. Moreover, any operator, satisfying the Kitai Criterion is hereditarily hypercyclic and therefore mixing [3]. Hypercyclicity in the proof of the above result in reference [27] is demonstrated via application of the Kitai Criterion. Thus the following slightly stronger statement holds.

**Corollary 13.6 *** Let . Then is hypercyclic if (13.1) is satisfied and is non-hypercyclic if (13.1) is not satisfied.*

Now we demonstrate that the answer to both parts of Question B^{2}CP is negative. Consider the subset U of being the interior of the triangle with vertices -1, I and -i. In other words Next, let . The boundary of V consists of the interval [−1+i,1+i] and two circle arcs. It is clear that U and V are bounded, open, connected and simply connected. From the definition of the sets U and V it immediately follows that the open set 1+U ={1+ z : z∈U} intersects the unit circle. On the other hand, since U is contained in the left half-plane, we see that Similarly, we see that and the open set e^{V} intersects the unit circle. According to the Riemann Theorem [48], there exist holomorphic homeomorphisms and . Obviously . Since and both and intersect the unit circle, Corollary 13.6 implies that and are hereditarily hypercyclic. Since , , is contained in , and does not intersect , Corollary 13.6 implies that and are non-hypercyclic. Finally, observe that is the closure of U and is the closure of -V and therefore the spectra of and are connected and contain 0. Taking into account that all separable infinite dimensional Hilbert spaces are isomorphic, we arrive to the following result, which answers negatively the Question B^{2}CP.

*Proposition 13.7** There exist bounded linear operators A, B on the complex Hilbert space _{2} such that σ(A) and σ(B) are connected and contain 0, I+A and e^{B} are hereditarily hypercyclic, while e^{A} and I+B are non-hypercyclic.*

Finally, if the answer to Question 13.4 is affirmative, then the following interesting question naturally arises.

**Problem 13.8** *Let A be a quasinilpotent bounded linear operator on a complex Banach space X and f be an entire function on one variable such that . Is it true that hypercyclicity of f(A) is equivalent to hypercyclicity of I+A?*

**Spaces C _{k}(M) and their duals**

Let (M,d) be a separable metric space and C_{k}(M) be the space of continuous functions with the compact-open topology (=the topology of uniform convergence on compact subsets of M). It is easy to see that C_{k}(M) is complete. Moreover, C_{k}(M) is metrizable if and only M is locally compact and C_{k}(M) carries weak topology if and only if M is discrete. On the other hand, there always is an _{1}- sequence with dense span inC_{k}(M). Indeed, let be a metrizable compactification of M. Since is a separable Banach space, there is an _{1}- sequence with dense span in . Since is densely and continuously embedded intoC_{k}(M), is an _{1}- sequence with dense span in C_{k}(M). Thus C_{k}(M) ∈M if and only if M is non-discrete. Corollary 1.4 implies now the following proposition.

**Proposition 13.9** *If (M,d) is a separable non-discrete metric space, then there is a hereditarily hypercyclic operator on C _{k}(X).*

The reason for inclusion of the above proposition is to demonstrate that Theorem 1.3 and Corollary 1.4 are applicable far beyond metrizable or LB-spaces. The spaces C_{k}(M) can have quite ugly structure indeed. For instance, take M being the set of rational numbers with the metric induced from , and you have got the space C_{k}(), which does not fall into any of the well-understood and studied classes of locally convex spaces. We would like to raise the following question.

**Problem 13.10** *Characterize separable metric spaces (M,d) such that C _{k}(M) supports a dual hypercyclic operator.*

It is worth noting that if M is discrete, then either C_{k}(M) is finite dimensional or is isomorphic to ω and therefore does not support a dual hypercyclic operator (there are no hypercyclic operators on φ=ω^{′}). In general, (M)=(C_{k} (M))′ can be naturally identified with the space of finite -valued Borel σ -additive measures on M with compact support. This dual space is separable in the strong topology if and only if all compact subsets of M are finite or countable. If it is not the case, there is no point to look for dual hypercyclic operators on C_{k}(M). Thus the only spaces (M,d) for which C_{k}(M) has a chance to support a dual hypercyclic operator are non-discrete spaces with no uncountable compact subsets. The first natural candidate to consider is .

Note also that although Theorem 1.15 provides answers to the questions of Petersson, mentioned in the introduction, it does not characterize Fréchet spaces, supporting a dual hypercyclic operator.

**Problem 13.11 ***Characterize Fréchet spaces X such that X supports a dual hypercyclic operator.*

The most natural Fréchet space for which we do not know whether it supports a dual hypercyclic operator is the countable power of the Hilbert space _{2}.

**The Hypercyclicity Criterion**

The following universality criterion is proved by Bés and Peris [2, Theorem 2.3 and Remark 2.6]. It is formulated in [2] in the case when X is an space, but the proof works without any changes for Baire separable metrizable topological vector spaces.

**Theorem BP** Let be a sequence of pairwise commuting continuous linear operators with dense range on a Baire separable metrizable topological vector space X. Then the following conditions are equivalent:

(a) The family is universal;

(b)There exists an infinite subset A of _{+} such that the familyis hereditarily universal

(c)There exist a strictly increasing sequence {n_{k}} of non-negative integers, dense subsets E and F of X and maps for k ∈ _{+ }such that , and as k→∞ for any x ∈ E and y ∈ F.

We formulate now the so-called Hypercyclicity and Supercyclicity Criteria, which follow easily from the above theorem.

**Theorem HC** *Let X be a Baire separable metrizable topological vector space and T∈ L(X). Then the following conditions are equivalent:*

(a)T ⊕ T is hypercyclic;

(b)There exists an infinite subset A of _{+} such that the family is hereditarily universal;

(c)There exist a strictly increasing sequence {n_{k}} of non-negative integers, dense subsets E and F of X and maps for k ∈ _{+ }such that ,, and as k→∞ for any x ∈ E and y ∈ F.

**Theorem SC.** *Let X be a Baire separable metrizable topological vector space and T∈ L(X). Then the following conditions are equivalent:*

*(a) T ⊕ T is supercyclic;*

*(b) There exists an infinite subset A of _{+} and a sequence of positive numbers such that the family is hereditarily universal;*

(c) There exist a strictly increasing sequence {n_{k}} of non-negative integers, dense subsets E and F of X, and a sequence * of positive numbers and maps for k ∈ _{+ }such that and , and as k→∞ for any x ∈ E and y ∈ F.*

An operator satisfying the condition (c) of Theorem HC (respectively Theorem SC) is said to satisfy the Hypercyclicity (respectively, Supercyclicity) Criterion. The long standing question whether any hypercyclic operator T on a Banach space satisfies the Hypercyclicity Criterion, was recently solved negatively by Read and De La Rosa [49]. Their result was extended by Bayart and Matheron [50], who demonstrated that on any separable Banach space with an unconditional Schauder basis such that the forward shift operator associated with this basis is bounded, there is a hypercyclic operator T such that T ⊕ T is not hypercyclic. This leaves open the following question raised in reference [50].

**Problem 13.12** *Does there exist a separable infinite dimensional Banach space X such that any hypercyclic operator on X satisfies the Hypercyclicity Criterion?*

It is observed in reference [50] that any T∈ L(ω) satisfies the Hypercyclicity Criterion. It also follows from Theorem 1.7 and Theorem HC. Thus the above question in the class of Fréchet spaces has an affirmative answer, which leads to the following problem.

**Problem 13.13** *Characterize separable infinite dimensional Fréchet spaces X on which any hypercyclic operator on X satisfies the Hypercyclicity Criterion?*

It is worth noting that non-hypercyclicity of T ⊕ T in references [49,50] is ensured by the existence of a non-zero continuous bilinear form b : X × X → with respect to which T is symmetric: b(Tx, y) = b(x,Ty) for any x, y∈X . The following proposition formalizes the corresponding implication. Similar statements have been proved by many authors in various particular cases. The proof goes along the same lines as in any of them.

Let *X* and *Y* be topological vector spaces and *b : X× X→Y* be separately continuous and bilinear. We say that *T∈ L(X)* is *b-symmetric if b(Tx, y) = b(x,Ty)* for any *x, y∈X* . Recall also that *b* is *called symmetric if b(x, y)=b(y, x)* for any* x, y∈X* and* b* is called *antisymmetric if b(x, y) = −b(y, x)* for any *x, y∈X .*

**Proposition 13.14** *Let X and Y be a topological vector spaces, b : X× X→Y be separately continuous, non-zero and bilinear and T∈ L(X) be b -symmetric. Then T ⊕ T is non-cyclic. If additionally b is nonsymmetric, then T2 is non-cyclic.*

**Proof. **Consider the left and right kernels of *b*:

(13.2)

Separate continuity of *b* implies that are closed linear subspaces of *X*. Since *b* is non-zero, we have *X _{0} ≠ X* and

Since and we have On the other hand, using *b* -symmetry of *T*, we have

Thus the orbit of (x, y) with respect to *T *⊕* T* lies in the proper closed linear subspace ker Φ of X, which contradicts cyclicity of (*x, y*) for *T* ⊕ *T*.

Assume now that b is non-symmetric. Then is a non-zero separately continuous bilinear map from *X × X *to* Y*. Moreover, *T* is c-symmetric. Assume that x is a cyclic vector for *T ^{2}*. Then

Hence the orbit of x with respect to *T ^{2}* lies in the proper closed linear subspace of

This looks like a proper place to reproduce the following question of Grivaux.

**Problem 13.15 ***Let X be a Banach space and T∈ L(X) be such that T ⊕ T is cyclic. Does it follow that T ^{2} is cyclic?*

As a straightforward toy illustration of the above proposition one can consider the following fact. Let be a measure space, , 0 < p < ∞ and be the operator of multiplication by *g : Tf = fg *for . Then *T⊕ T *is non-cyclic. Indeed, consider the continuous bilinear map . Clearly b is non-zero and *T* is* b*-symmetric. By Proposition 13.14,* T* ⊕ *T* is non-cyclic. The above mentioned result of Bayart and Matheron can now be formulated in the following way.

**Theorem BM** *Let X be a separable infinite dimensional Banach space with an unconditional Schauder basis such that the forward shift operator associated with this basis is bounded. Then there exists a hypercyclic T∈ L(X)and a non-zero continuous bilinear form such that T is b -symmetric. In particular, T *⊕* T is non-cyclic.*

The form *b* in the above theorem must be symmetric. Indeed, otherwise, by Proposition 13.14, *T ^{2}* is non-cyclic, which contradicts hypercyclicity of

**Problem 13.16** *Let T be a hypercyclic continuous linear operator on a Banach space X such that T *⊕* T is non-hypercyclic. Does there exist a non-zero symmetric continuous bilinear form such that T *⊕* T is b -symmetric?*

It is worth noting that non-existence of such a form *b* is equivalent to the density of the range of the operator* *projective tensor product * *Another observation concerning Theorem BM is that operators constructed in reference [50] have huge spectrum. Namely, their spectrum contains a disk centered at 0 of radius >1. On the other hand, we know (see Theorems 1.19 and 1.21) that any separable infinite dimensional complex Banach space supports plenty of hypercyclic operators with the spectrum being the singleton {1}. This leads to the following question.

**Problem 13.17** *Let T be a hypercyclic continuous linear operator on a complex Banach space X such that σ(T)= *{1}.* Is T *⊕* T hypercyclic?*

It is worth noting that an affirmative answer to the above question would take care of Problem 13.7. Indeed, the spectrum of any hypercyclic operator on a hereditarily indecomposable complex Banach space [51] is a singleton {z} with* z ∈*

**n-****supercyclic operators**

Recently Feldman [52] has introduced the notion of an n -supercyclic operator for n ∈ A bounded linear operator T on a Banach space *X* is called n-supercyclic for n ∈ if there exists an *n *-dimensional linear subspace *L* of *X* such that its orbit * *is dense in *X*. Such a space L is called an n - supercyclic subspace for *T*. Clearly,1- supercyclicity coincides with the usual supercyclicity. In reference [52], for any n ∈, n ≥ 2, a bounded linear operator *T* on is constructed, which is *n* -supercyclic and not (*n-1*)-supercyclic. The construction is based on the observation that if *T _{k}* for

**Proposition 13.18** *There exists a hypercyclic operator T∈ L() such that T *⊕* T is not n -supercyclic for any n ∈*

The above proposition follows immediately from Theorem BM and the next proposition, which implies that for the operator *T* from Theorem BM, *T* ⊕ *T* is not n -supercyclic for any n ∈.

**Proposition 13.19** *Let X be an infinite dimensional topological vector space, be non-zero, separately continuous and bilinear and T∈ L(X) be a b-symmetric operator with no non-trivial closed invariant subspaces of finite codimension. Then for any finite dimensional linear subspace L of X × X, the set is nowhere dense in X × X.*

In order to prove Proposition 13.19, we need the following lemma.

**Lemma 13.20** *Let m ∈ , L and X be topological vector spaces, such that and be a continuous bilinear map such that for each non-zero n ∈ L, the linear map is surjective. Then the set is closed and nowhere dense in X.*

**Proof.** First, observe that it is enough to prove the required statement in the case, when *X* is finite dimensional. Indeed, let e_{1,...,}e_{k} be a basis of *L* and Then *X _{1}* is a closed linear subspace such that for any . Moreover,

Thus without loss of generality, we can assume that *X* is finite dimensional. Consider the unit sphere *S* in *L* with respect to some Hilbert space norm on *L*. For each *u *∈* S*, is onto and we can pick an m-dimensional subspace *Z _{u}* of

It suffices to show that each Aj is closed and nowhere dense. Let Closeness of *A _{j}* is rather easy. Indeed, let be a sequence of elements of

**Proof of Proposition 13.9. **First, we consider the case of nondegenerate *b*. That is, we assume that both the left and the right kernels *X _{0}* and

First, we shall check that for any (*x, y*) ≠ (0,0) , the functionals* Φk (x, y) for k ∈ *are linearly independent. Assume the contrary. Then there exists a non-zero polynomial p such that *b(p(T)x,v)=b(u, p(T)y)* for any *u,v* ∈ *X*. Since the left-hand side of the last equality does not depend on u and the right-hand side does not depend on v, they both do not depend on both u and v. Hence *b(p(T)x,v) = b(u, p(T)y) = 0* for any* u,v *∈* X. *Since *T* is *b* -symmetric, we have *b(x, p(T)v)=b(p(T)u, y)*= 0 for any *u,v* ∈ *X*. Hence *,* where *,.*Since *T* has no non-trivial closed invariant subspaces of finite codimension, *T'* has no non-trivial finite dimensional invariant subspaces. By Lemma 7.2, *p*(*T'* ) is injective. Hence .Since *b* is non-degenerate, we then have *x=y=0*, which contradicts with the assumption . Thus the functionals for k ∈ are linearly independent for each *.*

Let *L* be a finite dimensional linear subspace of *X × X, and m ∈ , m > n. *Consider the bilinear map

Since for any non-zero ,the functionals* *are linearly independent, we see that the linear map is onto. By Lemma 13.20, the set

is closed and nowhere dense in *X *× *X*. Let now for some p ∈Then for some * *Then for any *j *∈since *T* is *b*-symmetric. It follows that (*u,v*)∈*A* . That is, the nowhere dense set *A* contains the set *.* Thus the latter set is nowhere dense.

It remains to reduce the general case to the case of nondegenerate *b*. Since *b* is non-zero, at least one of the bilinear forms or is non-zero. Clearly *T* is symmetric with respect to both *b _{0}* and

The following question is raised by Bourdon, Feldman and Shapiro [53].

**Problem 13.21 ***Let X be a complex Banach space, n ∈ and T ∈ L(X) be such that T is n-supercyclic and Is T cyclic?*

It is worth noting that the only known examples of n -supercyclic operators *T* with are the mentioned direct sums of operators satisfying the Supercyclicity Criterion with the same sequence {n_{k}}. Such direct sums are all cyclic [54].

Feldman in reference [52] has also introduced the concept of an ∞-supercyclic operator. A bounded linear operator *T* on a Banach space *X* is called ∞ -supercyclic if there exists a linear subspace *L* of *X* such that its orbit {*T ^{n} x: n ∈ , t ∈ L*} is dense in

is not supercyclic. In [56] it is shown that V is not n -supercyclic for any* n ∈ .*However, it turns out that *V* is ∞-supercyclic.

**Proposition 13.22** *The Volterra operator is ∞-supercyclic.*

**Proof.** For any non-zero *h* ∈ *L _{2}*[0, 1] we denote by

* *(13.3)

is the adjoint of *V*. Consider the space If h is non-zero element of *ε*, then according to the above display, . Thus by (13.3)

(13.4)

Consider now the following specific h ∈ ε : h(1) =0 * and *for 0 ≤ x <1 . First, we shall show that *L _{h}* does not contain any non-zero invariant subspace of

On the other hand, analyticity of h on [0, *q*] and easy lower estimates if * *imply that

From the last two displays it follows that as *n* → ∞ . Hence contains the space of all functions vanishing on a neighborhood of 1. Since the latter space is dense in , A is dense in , which completes the proof.

- **cyclicity and supercyclicity**

Let *T* be a continuous linear operator on a separable complex topological vector space *X*. We say that *T* is - *cyclic* if there exists *x *∈ *X* such that the linear span of the orbit {*T ^{n}*

**Theorem LM** *Let T be a continuous linear operator on a complex locally convex space X with . Then T is supercyclic if and only if T is ^{+}-supercyclic.*

As we have shown, there are bilateral weighted shifts *T _{w}* on () with the weight sequence

**Proposition 13.23 ***Let be a weight such that for any n ∈ , and let p be a polynomial with real coefficients. Then the operator p(T _{w}) acting on complex () is not -cyclic. In particular, by Theorem LM, p(T_{w}) is not weakly supercyclic.*

**Corollary 13.24 ***Let with for any n ∈ Then the operator I+T _{w} acting on () is not weakly supercyclic.*

**Proof of Proposition 13.23. **First, note that if * *satisfy for any *n ∈ *, then *T _{w}* and

Since () for 1 ≤ *p* ≤ 2 is contained in () and carries a stronger topology than the one inherited from (), Proposition 13.23 remains true if we replace () by () with 1 ≤ p ≤ 2 . On the other hand, the unweighted shift on () with 2< *p* < ∞ is weakly supercyclic [37] and therefore the statement of Proposition 13.23 becomes false if we replace () by () with *p* > 2. At this point it is interesting to remind that, according to Theorem B, norm hypercyclicity and supercyclicity of a bilateral weighted shift on () do not depend on *p*. This leads naturally to the following question.

**Problem 13.25** *Characterize hypercyclicity and supercyclicity of the operators of the form I*+*T, where T is a bilateral weighted shift on (). In particular, does hypercyclicity or supercyclicity of these operators depend on the choice of P. *1* ≤ p < +*∞* ?*

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